23.26 - Properties of Definite Integrals

In the previous section, we saw evaluation of definite integrals by substitution. We saw some solved examples also. In this section, we will see some properties of definite integrals.

$\bf{{\rm{P_0}}:\int_a^b{\left[f(x) \right]dx}~=~\int_a^b{\left[f(t) \right]dt}}$

Proof:

The function is the same. Upper and lower limits are also the same. We are changing only the variable. So the definite integral will be the same.

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$\bf{{\rm{P_1}}:\int_a^b{\left[f(x) \right]dx}~=~-\int_b^a{\left[f(x) \right]dx}}$

Proof can be written in 4 steps:

1. $\small{\int_a^b{\left[f(x) \right]dx}~=~F(b)\,-\,F(a)}$

2. $\small{\int_b^a{\left[f(x) \right]dx}~=~F(a)\,-\,F(b)~=~-\left[F(b)\,-\,F(a) \right]}$

3. From (2), we get:
$\small{-\int_b^a{\left[f(x) \right]dx}~=~F(b)\,-\,F(a) }$

4. Comparing the results in (1) and (3), we get:
$\small{\int_a^b{\left[f(x) \right]dx}~=~-\int_b^a{\left[f(t) \right]dx}}$

5. Note that:
$\small{\int_a^a{\left[f(x) \right]dx}~=~F(a)\,-\,F(a)~=~0}$

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$\bf{{\rm{P_2}}:\int_a^b{\left[f(x) \right]dx}~=~\int_a^c{\left[f(x) \right]dx}~+~\int_c^b{\left[f(x) \right]dx}}$

Proof can be written in 3 steps:

1. $\small{\int_a^c{\left[f(x) \right]dx}~=~F(c)\,-\,F(a)}$

2. $\small{\int_c^b{\left[f(x) \right]dx}~=~F(b)\,-\,F(c)}$

3. Therefore:

$\small{\int_a^c{\left[f(x) \right]dx}~+~\int_c^b{\left[f(x) \right]dx}}$

$\small{~=~F(c)\,-\,F(a)~+~F(b)\,-\,F(c)}$

$\small{~=~F(b)\,-\,F(a)}$

$\small{~=~\int_a^b{\left[f(x) \right]dx}}$

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$\bf{{\rm{P_3}}:\int_a^b{\left[f(x) \right]dx}~=~\int_a^b{\left[f(a+b-x) \right]dx}}$

Proof can be written in 3 steps:

1. Put $\small{t=a+b-x}$

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

2. Rearranging the limits:

   ♦ When x approach the lower limit 'a', t approach 'b'

   ♦ When x approach the upper limit 'b', t approach 'c'
   
3. So we can write:

$\small{\int_a^b{\left[f(a+b-x) \right]dx}~=~\int_a^b{\left[(-1)(-1)f(a+b-x) \right]dx}}$

$\small{~=~\int_b^a{\left[(-1)\,f(t) \right]dt}~=~(-1)\int_b^a{\left[\,f(t) \right]dt}}$

$\small{~=~\int_a^b{\left[\,f(t) \right]dt}}$ (by P₁)

$\small{~=~\int_a^b{\left[\,f(x) \right]dx}}$ (by P0q) 

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$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

Proof can be written in 4 steps:

1. Put $\small{t=a-x}$

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

2. Rearranging the limits:

   ♦ When x approach the lower limit zero, t approach 'a'

   ♦ When x approach the upper limit 'a', t approach zero
   
3. So we can write:

$\small{\int_a^b{\left[f(a-x) \right]dx}~=~\int_a^b{\left[(-1)(-1)f(a-x) \right]dx}}$

$\small{~=~\int_a^0{\left[(-1)\,f(t) \right]dt}~=~(-1)\int_a^0{\left[\,f(t) \right]dt}}$

$\small{~=~\int_0^a{\left[\,f(t) \right]dt}}$ (by P₁)

$\small{~=~\int_0^a {\left[\,f(x) \right]dx}}$ (by P0q)

4. Note that, P₄ is a particular case of P₃.

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$\bf{{\rm{P_5}}:\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a{\left[f(2a-x) \right]dx}}$

Proof can be written in 4 steps:

1. Using P₂, we can write:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_a^{2a}{\left[f(x) \right]dx}}$

• We can denote this as:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~I_1~+~I_2}$

2. Put $\small{t=2a-x}$

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

Also, $\small{x = 2a - t}$

3. Rearranging the limits of I₂:

   ♦ When x approach the lower limit 'a', t approach 'a'

   ♦ When x approach the upper limit '2a', t approach zero
   
So I₂ can be written as:

$\small{\int_a^{2a}{\left[f(x) \right]dx}~=~\int_a^0{\left[f(2a-t) \right]dx}~=~\int_a^0{\left[(-1)(-1)f(2a-t) \right]dx}}$

$\small{~=~\int_a^0{\left[(-1)\,f(2a - t) \right]dt}~=~(-1)\int_a^0{\left[\,f(2a - t) \right]dt}}$

$\small{~=~\int_0^a{\left[\,f(2a - t) \right]dt}}$ (by P₁)

$\small{~=~\int_0^a {\left[\,f(2a - x) \right]dx}}$ (by P0q)

4. So from step (1), we get:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(2a - x) \right]dx}}$

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$\bf{{\rm{P_6}}:}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f(2a - x)~=~f(x)}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f(2a - x)~=~-f(x)}$

Proof can be written in 3 steps:

1. Using P₅, we can write:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(2a - x) \right]dx}}$

2. If $\small{f(2a-x)~=~f(x)}$, (1) becomes:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a{\left[f(x) \right]dx}~=~2\int_0^a{\left[f(x) \right]dx}}$

3. If $\small{f(2a-x)~=~-f(x)}$, (1) becomes:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~-~\int_0^a{\left[f(x) \right]dx}~=~0}$

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$\bf{{\rm{P_7}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f}$ is an even function
         ✰ That is., if $\small{f(-x)~=~f(x)}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f}$ is an odd function
         ✰ That is., if $\small{f(-x)~=~-f(x)}$

Proof can be written in 7 steps:

1. Using P₂, we can write:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~\int_{-a}^0{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}}$

• We can denote this as:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~I_1~+~I_2}$

2. Put $\small{t=-x}$ in I₁

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

Also, $\small{x = - t}$

3. Rearranging the limits of I₁:

   ♦ When x approach the lower limit '-a', t approach 'a'

   ♦ When x approach the upper limit zero, t approach zero
   
4. So I₁ can be written as:

$\small{\int_{-a}^0{\left[f(x) \right]dx}~=~\int_{a}^0{\left[f(-t) \right]dx}~=~\int_{a}^0{\left[(-1)(-1)f(-t) \right]dx}}$

$\small{~=~\int_{a}^0{\left[(-1)\,f(- t) \right]dt}~=~(-1)\int_{a}^0{\left[f(-t) \right]dt}}$

$\small{~=~(-1)\int_{a}^0{\left[f(-x) \right]dx}}$ (by P0q)

$\small{~=~\int_{0}^{a}{\left[f(-x) \right]dx}}$ (by P₁)

5. So from step (1), we get:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~\int_{0}^a{\left[f(-x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}}$

6. If $\small{f}$ is an even function, then $\small{f(-x)~=~f(x)}$

• In such a situation, step (5) becomes:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~\int_{0}^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}~=~2\int_0^a {\left[\,f(x) \right]dx}}$

7. If $\small{f}$ is an odd function, then $\small{f(-x)~=~-f(x)}$

• In such a situation, step (5) becomes:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~-\int_{0}^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}~=~0}$

◼ We can cite two examples related to P₇:

Example 1:
This can be written in 5 steps:

1. $\small{f(x) \,=\,x^2}$ is an even function.

• Suppose that, we need to evaluate $\small{\int_{-1}^{1}{\left[x^2 \right]dx}}$

2. The usual approach is:

$\small{\int_{-1}^{1}{\left[x^2 \right]dx}\,=\,\left[\frac{x^3}{3} \right]_{-1}^1\,=\,\frac{1}{3}\,-\,\left(\frac{-1}{3} \right)\,=\,\frac{2}{3}}$

3. But by applying P₇, we can straight away write:

$\small{\int_{-1}^{1}{\left[x^2 \right]dx}\,=\,2\int_{0}^{1}{\left[x^2 \right]dx}}$

$\small{\,=\,2 \left[\frac{x^3}{3} \right]_{0}^1\,=\,2\left[\frac{1}{3}\,-\,\frac{0}{3} \right]\,=\,\frac{2}{3}}$

4. We get the same result in both (2) and (3).

5. In the fig.23.24 below,

   ♦ the area of the blue region from x = -1 to x = 0 is 1/3

   ♦ the area of the blue region from x = 0 to x = 1 is also 1/3

Fig.23.24

 

 

• Both areas are positive. So the total area from x = -1 to x = 1 will be twice (1/3)

Example 2:
This can be written in 5 steps:

1. $\small{f(x) \,=\,x^3}$ is an odd function.

• Suppose that, we need to evaluate $\small{\int_{-1}^{1}{\left[x^3 \right]dx}}$

2. The usual approach is:

$\small{\int_{-1}^{1}{\left[x^3 \right]dx}\,=\,\left[\frac{x^4}{4} \right]_{-1}^1\,=\,\frac{1}{4}\,-\,\frac{1}{4}\,=\,0}$

3. But by applying P₇, we can straight away write:

$\small{\int_{-1}^{1}{\left[x^3 \right]dx}\,=\,0}$

4. We get the same result in both (2) and (3).

5. In the fig.23.25 below,

   ♦ the area of the blue region from x = -1 to x = 0 is 1/4

   ♦ the area of the blue region from x = 0 to x = 1 is also 1/4

Fig.23.25

• The area from x = -1 to x = 0 is -ve. But the area from x = 0 to x = 1 is +ve. So the total area from x = -1 to x = 1 will be zero.

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Now we will see some solved examples:

Solved Example 23.96
Given that:
$\small{\int_{1}^{5}{\left[f(x) \right]dx}\,=\,-3}$
$\small{\int_{2}^{5}{\left[f(x) \right]dx}\,=\,4}$

Find:
$\small{\int_{1}^{2}{\left[f(x) \right]dx}}$
Solution:
1. Applying P₂, we can write:

$\small{\int_{1}^{5}{\left[f(x) \right]dx}\,=\,\int_{1}^{2}{\left[f(x) \right]dx}\,+\,\int_{2}^{5}{\left[f(x) \right]dx}}$

2. Substituting the known values, we get:

$\small{-3\,=\,\int_{1}^{2}{\left[f(x) \right]dx}\,+\,4}$

3. Therefore,

$\small{\int_{1}^{2}{\left[f(x) \right]dx}\,=\,-3 - 4  = -7}$

Solved Example 23.97

Evaluate $\small{\int_{-2}^{3}{\left[|x| \right]dx}}$
Solution:
1. We know that:
$\left|x \right| = \begin{cases} x,  & \text{if}~x \ge 0 \\[1.5ex] -x, & \text{if}~x<0 \end{cases}$

2. The integrand changes at zero. So we will split the interval at zero. Then by applying P₂, it can be written as:

$\small{\int_{-2}^{3}{\left[\left|x \right| \right]dx}\,=\,\int_{-2}^{0}{\left[\left|x \right| \right]dx}\,+\,\int_{0}^{3}{\left[\left|x \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-2}^{3}{\left[\left|x \right| \right]dx}\,=\,I_1\,+\,I_2}$

3. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x \right|\,=\,-x,~\text{if}~x<0}$

• For I₂, we must use the segment: $\small{\left|x \right|\,=\,x,~\text{if}~x \ge 0}$

4. So from (2), we get:

$\small{\int_{-2}^{3}{\left[\left|x \right| \right]dx}\,=\,\int_{-2}^{0}{\left[-x \right]dx}\,+\,\int_{0}^{3}{\left[x \right]dx}}$

$\small{\,=\,(-1)\int_{-2}^{0}{\left[x \right]dx}\,+\,\int_{0}^{3}{\left[x \right]dx}}$

$\small{\,=\,(-1)\left[\frac{x^2}{2} \right]_{-2}^{0}\,+\,\left[\frac{x^2}{2} \right]_{0}^{3}}$

$\small{\,=\,(-1)\left[0\,-\,\frac{(-2)^2}{2} \right]\,+\,\left[\frac{3^2}{2}\,-\,0 \right]}$

$\small{\,=\,(-1)\left[-2 \right]\,+\,\left[\frac{9}{2} \right]~=~\frac{13}{2}}$

Solved Example 23.98

Evaluate $\small{\int_{-1}^{1}{\left[1\,-\,|x| \right]dx}}$
Solution:
1. We know that:
$\left|x \right| = \begin{cases} x,  & \text{if}~x \ge 0 \\[1.5ex] -x, & \text{if}~x<0 \end{cases}$

2. The integrand changes at zero. So we will split the interval at zero. Then by applying P₂, it can be written as:

$\small{\int_{-1}^{1}{\left[1\,-\,\left|x \right| \right]dx}\,=\,\int_{-1}^{0}{\left[1\,-\,\left|x \right| \right]dx}\,+\,\int_{0}^{1}{\left[1\,-\,\left|x \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-1}^{1}{\left[1\,-\,\left|x \right| \right]dx}\,=\,I_1\,+\,I_2}$

3. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x \right|\,=\,-x,~\text{if}~x<0}$

• For I₂, we must use the segment: $\small{\left|x \right|\,=\,x,~\text{if}~x \ge 0}$

4. So from (2), we get:

$\small{\int_{-1}^{1}{\left[1\,-\,\left|x \right| \right]dx}\,=\,\int_{-1}^{0}{\left[1+x \right]dx}\,+\,\int_{0}^{1}{\left[1\,-\,x \right]dx}}$

$\small{\,=\,\left[x\,+\,\frac{x^2}{2} \right]_{-1}^{0}\,+\,\left[x\,-\,\frac{x^2}{2} \right]_{0}^{1}}$

$\small{\,=\,\left[0\,+\,\frac{0^2}{2}\,-\,\left(-1\,+\,\frac{(-1)^2}{2} \right) \right]\,+\,\left[1\,-\,\frac{1^2}{2}\,-\,\left(0\,-\,\frac{0^2}{2} \right) \right]}$

$\small{\,=\,\left[-\,\left(-1\,+\,\frac{1}{2} \right) \right]\,+\,\left[1\,-\,\frac{1}{2}\,-\,0 \right]}$

$\small{\,=\,\left[1\,-\,\frac{1}{2} \right]\,+\,\left[1\,-\,\frac{1}{2}\right]}$

$\small{\,=\,\left[\frac{1}{2} \right]\,+\,\left[\frac{1}{2}\right]\,=\,1}$

Solved Example 23.99
Evaluate $\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}}$
Solution:
1. Let us determine the intervals where (x−1) is +ve or −ve. For that, first we solve the inequality: x−1<0

We have: $\small{x-1 < 0}$

$\small{\Rightarrow x < 1}$

• So when x is less than 1, (x−1) will be -ve.

• That means, when x is less than 1, $\small{\left|x-1 \right|\,=\,-(x-1)}$

2. Next we solve the inequality:x−1>0

We have: $\small{x-1 > 0}$

$\small{\Rightarrow x > 1}$

• So when x is greater than 1, (x-1) will be +ve.

• That means, when x is greater  than 1, $\small{\left|x-1 \right|\,=\,x-1}$

3. Also, we must solve the equation x-1 = 0

This gives x = 1

• So when x is equal to 1, (x-1) will be zero.

• That means, when x is equal to 1, $\small{\left|x-1 \right|\,=\,x-1}$

4. Now we can write a piece wise function:

$\left|x-1 \right| = \begin{cases} x-1,  & \text{if}~x \ge 1 \\[1.5ex] -(x-1), & \text{if}~x<1  \end{cases}$

5. The integrand changes at 1. So we will split the interval at 1. Then by applying P₂, it can be written as:

$\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}\,=\,\int_{0}^{1}{\left[\left|x-1 \right| \right]dx}\,+\,\int_{1}^{4}{\left[\left|x-1 \right| \right]dx}}$

• We will denote it as:

$\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}\,=\,I_1\,+\,I_2}$

6. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x-1 \right|\,=\,-(x-1),~\text{if}~x<1}$

• For I₂, we must use the segment: $\small{\left|x-1 \right|\,=\,x-1,~\text{if}~x \ge 1}$

7. So from (5), we get:

$\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}\,=\,\int_{0}^{1}{\left[-(x-1) \right]dx}\,+\,\int_{1}^{4}{\left[x-1 \right]dx}}$

$\small{\,=\,(-1)\int_{0}^{1}{\left[x-1 \right]dx}\,+\,\int_{1}^{4}{\left[x-1 \right]dx}}$

$\small{\,=\,(-1)\left[\frac{x^2}{2}\,-\,x \right]_{0}^{1}\,+\,\left[\frac{x^2}{2}\,-\,x \right]_{1}^{4}}$

$\small{\,=\,(-1)\left[\frac{1^2}{2}\,-\,1\,-\,\left(\frac{0^2}{2}\,-\,0 \right) \right]\,+\,\left[\frac{4^2}{2}\,-\,4\,-\,\left(\frac{1^2}{2}\,-\,1 \right) \right]}$

$\small{\,=\,(-1)\left[\frac{1}{2}\,-\,1\,-\,\left(0 \right) \right]\,+\,\left[\frac{16}{2}\,-\,4\,-\,\left(\frac{1}{2}\,-\,1 \right) \right]}$

$\small{\,=\,(-1)\left[-\frac{1}{2} \right]\,+\,\left[4\,+\,\frac{1}{2}\right]\,=\,5}$

Solved Example 23.100
Evaluate $\small{\int_{-1}^{2}{\left[\left|x^3\,-\,x \right| \right]dx}}$
Solution:
1. Let us determine the intervals where (x³ - x) is +ve or -ve.

2. For that, first we need to solve the equation: $\small{f(x)=x^3 - x = x(x^2 - 1)=0}$

• The solution is: x = −1, x = 0 and x = 1

• So the given interval [−1,2] can be split into three intervals: [−1,0], [0,1] and [1,2]  

3. Consider the interval (−1,0). In this interval, all x values are -ve fractions. So $\small{x^2}$ will be a +ve fraction. Consequently, $\small{(x^2 - 1)}$ will be -ve and $\small{x(x^2 - 1)}$ will be +ve

So for this interval, we can write:

$\small{\left|x(x^2 - 1) \right|=x(x^2 - 1)~\text{if}~-1 < x < 0}$

4. Consider the interval (0,1). In this interval, all x values are +ve fractions. So $\small{x^2}$ will be a +ve fraction. Consequently, $\small{(x^2 - 1)}$ will be -ve and $\small{x(x^2 - 1)}$ will be -ve

So for this interval, we can write:

$\small{\left|x(x^2 - 1) \right|=-x(x^2 - 1)~\text{if}~0 < x < 1}$

5. Consider the interval (1,2). In this interval, all x values are greater than 1. So $\small{x^2}$ will be greater than 1. Consequently, $\small{(x^2 - 1)}$ will be +ve and $\small{x(x^2 - 1)}$ will be +ve.

So for this interval, we can write:

$\small{\left|x(x^2 - 1) \right|=x(x^2 - 1)~\text{if}~1 < x < 2}$

6. Consider the exact points x = −1, x = 0, x = 1 and x = 2

• When x = −1, $\small{x(x^2 - 1)} = 0$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= \pm x(x^2 - 1)~\text{if}~ x = -1}$

• When x = 0, $\small{x(x^2 - 1)} = 0$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= \pm x(x^2 - 1)~\text{if}~ x = 0}$

• When x = 1, $\small{x(x^2 - 1)} = 0$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= \pm x(x^2 - 1)~\text{if}~ x = 1}$

• When x = 2, $\small{x(x^2 - 1)} = 6$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= +x(x^2 - 1)~\text{if}~ x = 2}$

7. Combining (3), (4), (5) and (6), we can write:

$\left|x(x^2 - 1) \right| = \begin{cases} x(x^2 - 1),  & \text{if}~-1 \le x \le 0 \\[1.5ex] -x(x^2 - 1), & \text{if}~~0 \le x \le 1 \\[1.5ex] x(x^2 - 1), & \text{if}~~1 \le x \le 2  \end{cases}$

8. The integrand changes at −1, 0 and 1. So we will split the interval at those points. Then by applying P₂, it can be written as:

$\small{\int_{-1}^{2}{\left[\left|x^3-x \right| \right]dx}\,=\,\int_{-1}^{2}{\left[\left|x(x^2-1) \right| \right]dx}}$

$\small{\,=\,\int_{-1}^{0}{\left[\left|x(x^2-1) \right| \right]dx}\,+\,\int_{0}^{1}{\left[\left|x(x^2-1) \right| \right]dx}\,+\,\int_{1}^{2}{\left[\left|x(x^2-1) \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-1}^{2}{\left[\left|x(x^2-1) \right| \right]dx}\,=\,I_1\,+\,I_2\,+\,I_3}$

9. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x(x^2-1) \right|\,=\,x(x^2-1),~\text{if}~-1 \le x \le 0}$

• For I₂, we must use the segment: $\small{\left|x(x^2-1) \right|\,=\,-x(x^2-1),~\text{if}~0 \le x \le 1}$

• For I₃, we must use the segment: $\small{\left|x(x^2-1) \right|\,=\,x(x^2-1),~\text{if}~1 \le x \le 2}$

10. So from (8), we get:

$\small{\int_{-1}^{2}{\left[\left|x(x^2-1) \right| \right]dx}=}$

$\small{\,=\,\int_{-1}^{0}{\left[x(x^2-1) \right]dx}\,+\,\int_{0}^{1}{\left[-x(x^2-1) \right]dx}\,+\,\int_{1}^{2}{\left[x(x^2-1) \right]dx}}$

$\small{\,=\,\int_{-1}^{0}{\left[x^3-x \right]dx}\,+\,(-1)\int_{0}^{1}{\left[x^3-x \right]dx}\,+\,\int_{1}^{2}{\left[x^3-x \right]dx}}$

$\small{\,=\,\left[\frac{x^4}{4}\,-\,\frac{x^2}{2} \right]_{-1}^{0}\,+\,(-1) \left[\frac{x^4}{4}\,-\,\frac{x^2}{2} \right]_{0}^{1}\,+\, \left[\frac{x^4}{4}\,-\,\frac{x^2}{2} \right]_{1}^{2}}$

$\small{\,=\,\left[\frac{0^4}{4}\,-\,\frac{0^2}{2}\,-\,\left(\frac{(-1)^4}{4}\,-\,\frac{(-1)^2}{2} \right) \right]\,+\,(-1)\left[\frac{1^4}{4}\,-\,\frac{1^2}{2}\,-\,\left(\frac{0^4}{4}\,-\,\frac{0^2}{2} \right) \right]}$

$\small{~~~~~~+\left[\frac{2^4}{4}\,-\,\frac{2^2}{2}\,-\,\left(\frac{1^4}{4}\,-\,\frac{1^2}{2} \right) \right]}$

$\small{\,=\,\left[0\,-\,\left(\frac{-1}{4} \right) \right]\,+\,(-1)\left[\frac{-1}{4}\,-\,\left(0 \right) \right]+\left[2\,-\,\left(\frac{-1}{4} \right) \right]}$

$\small{\,=\,\left[\frac{1}{4} \right]\,+\,\left[\frac{1}{4} \right]+\left[\frac{9}{4} \right]\,=\,\frac{11}{4}}$

Solved Example 23.101
Evaluate $\small{\int_{-1}^{0}{\left[\left|4x\,+\,3 \right| \right]dx}}$
Solution:
1. Let us determine the intervals where (4x + 3) is +ve or -ve.

2. For that, first we need to solve the inequality: $\small{4x + 3 < 0}$

• The solution can be obtained as follows:
$\small{4x + 3 - 3  < 0 - 3}$
$\small{\Rightarrow 4x  <  - 3}$
$\small{\Rightarrow x  <  - \frac{3}{4}}$

That means, when x is less than $\small{- \frac{3}{4}}$, (4x+3) will be less than zero.  

3. Next  we need to solve the inequality: $\small{4x + 3 > 0}$

• The solution can be obtained as follows:
$\small{4x + 3 - 3  > 0 - 3}$
$\small{\Rightarrow 4x  >  - 3}$
$\small{\Rightarrow x  >  - \frac{3}{4}}$

That means, when x is greater than $\small{- \frac{3}{4}}$, (4x+3) will be greater than zero.

4. We have:

 $\small{\frac{-3}{4}\,=\,-0.75}$

• So the given interval [−1,0] can be split into two intervals:

$\small{\left(-1,-0.75 \right),~\left(-0.75,0 \right)}$

5. Consider the interval (−1, −0.75)
Based on (2), we can write:
(4x+3) will be −ve in this interval

So for this interval, we can write:

$\small{\left|4x + 3 \right|= -(4x + 3)~\text{if}~-1 < x < -0.75}$

6. Consider the interval (-0.75,0)
Based on (3), we can write:
(4x+3) will be +ve in this interval

So for this interval, we can write:
$\small{\left|4x + 3 \right|= (4x + 3)~\text{if}~-0.75 < x < 0}$

7. Consider the exact points x = −1, x = −0.75, and x = 0

• When x = −1, $\small{4x + 3} = -1$

So for this point, we can write:

$\small{\left|4x + 3 \right|= -(4x + 3)~\text{if}~ x = -1}$

• When x = -0.75, $\small{4x + 3} = 0$

So for this point, we can write:

$\small{\left|4x + 3 \right|= \pm(4x + 3)~\text{if}~ x = -0.75}$

• When x = 0, $\small{4x + 3} = 3$

So for this point, we can write:

$\small{\left|4x + 3 \right|= 4x + 3~\text{if}~ x = 0}$

8. Combining (3), (4) and (5), we can write:

$\left|4x + 3 \right| = \begin{cases} -(4x + 3),  & \text{if}~-1 \le x \le -0.75  \\[1.5ex] 4x + 3, & \text{if}~~-0.75 \le x \le 0  \end{cases}$

9. The integrand changes at -0.75. So we will split the interval at that point. Then by applying P₂, it can be written as:

$\small{\int_{-1}^{0}{\left[\left|4x+3 \right| \right]dx}\,=\,\int_{-1}^{-0.75}{\left[\left|4x+3 \right| \right]dx}\,+\,\int_{-0.75}^{0}{\left[\left|4x+3 \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-1}^{0}{\left[\left|4x+3 \right| \right]dx}\,=\,I_1\,+\,I_2}$

10. Choosing the appropriate segments:

• For I₁, we must use the segment:

$\small{\left|4x + 3 \right|\,=\,-(4x+3),~\text{if}~-1 \le x \le -0.75}$

• For I₂, we must use the segment:

$\small{\left|4x + 3 \right|\,=\,4x+3,~\text{if}~-0.75 \le x \le 0}$

11. So from (9), we get:

$\small{\int_{-1}^{0}{\left[\left|4x+3 \right| \right]dx}=}$

$\small{\,=\,\int_{-1}^{-0.75}{\left[-(4x+3) \right]dx}\,+\,\int_{-0.75}^{1}{\left[4x+3 \right]dx}}$

$\small{\,=\,(-1)\int_{-1}^{-0.75}{\left[4x+3 \right]dx}\,+\,\int_{-0.75}^{1}{\left[4x+3 \right]dx}}$

$\small{\,=\,(-1)\left[\frac{4 x^2}{2}\,+\,3x \right]_{-1}^{-0.75}\,+\,\left[\frac{4 x^2}{2}\,+\,3x \right]_{-0.75}^{0}}$

$\small{\,=\,(-1)\left[2x^2\,+\,3x \right]_{-1}^{-0.75}\,+\,\left[2x^2\,+\,3x \right]_{-0.75}^{0}}$

$\small{\,=\,(-1)\left[2(-0.75)^2\,+\,3(-0.75)\,-\,\left(2(-1)^2\,+\,3(-1) \right) \right]}$

$\small{~~~~~~+\left[2(0)^2\,+\,3(0)\,-\,\left(2(-0.75)^2\,+\,3(-0.75) \right) \right]}$

$\small{\,=\,\frac{5}{4}}$

---

In the next section, we will see a few more solved examples.

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23.25 - Evaluation of Definite Integrals by Substitution

In the previous section, we saw how the fundamental theorem of calculus can be used to find the definite integral. We saw some solved examples also. In this section, we will see how the method of substitution can be used to speed up the process. We will demonstrate the method using a solved example.

Solved Example 23.88
Evaluate the definite integral $\small{\int_{1}^{2}{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^{1}{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$

(a) Put $\small{u = (x^5 + 1)~\Rightarrow \frac{du}{dx} = 5x^4 \Rightarrow 5x^4 dx = du}$

(b) $\small{F~=~\int{\left[5x^4 \sqrt{x^5 + 1} \right]dx}~=~\int{\left[\sqrt{u} \right]du}}$

$\small{~=~\frac{u^{3/2}}{3/2}~=~\frac{2}{3} \left(x^5 + 1 \right)^{\frac{3}{2}}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(-1)~=~\left[\frac{2}{3} \left(x^5 + 1 \right)^{\frac{3}{2}} \right]_{-1}^{1}}$

$\small{~=~\left[\frac{2}{3} \left(1^5 + 1 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left((-1)^5 + 1 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\left[\frac{2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left(0 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\left[\frac{4 \sqrt 2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[0 \right]~=~\frac{4 \sqrt 2}{3}}$

• The above two steps are already familiar to us. Let us see a method to speed up the process. For that, we will analyze the upper and lower limits.

• We wrote: $\small{u = x^5 + 1}$.
So when x gets closer and closer to -1, u gets closer and closer to zero.
Also, when x gets closer and closer to 1, u gets closer and closer to 2.

• So when using u as the variable, the lower and upper limits are zero and 2 respectively.

• We wrote: $\small{F~=~\int{\left[\sqrt{u} \right]du}~=~\frac{u^{3/2}}{3/2}~=~\frac{2}{3}u^{\frac{3}{2}}}$

• So we can write:

$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{2}{3}u^{\frac{3}{2}} \right]_{0}^{2}}$

$\small{~=~\left[\frac{2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left(0 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\frac{4 \sqrt 2}{3}}$

---

Let us see a few more solved examples:

Solved Example 23.89
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{x}{x^2+1} \right]dx}}$

Put $\small{u~=~x^2 + 1 \Rightarrow \frac{du}{dx}~=~2x \Rightarrow 2x\,dx~=~du}$

So we have: $\small{F~=~\int{\left[\frac{2x}{2(x^2+1)} \right]dx}~=~\int{\left[\frac{1}{2(u)} \right]du}~=~\frac{\log \left|u \right|}{2}}$

2. We wrote: $\small{u~=~x^2 + 1}$

   ♦ When x approach the lower limit zero, u approach 1

   ♦ When x approach the upper limit 1, u approach 2

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}~=~\int_1^2{\left[\frac{1}{2(u)} \right]du}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(1)~=~\left[\frac{\log \left|u \right|}{2} \right]_{1}^{2}}$

$\small{~=~\left[\frac{\log \left(2 \right)}{2} \right]~-~\left[\frac{\log \left(1 \right)}{2} \right]~=~\left[\frac{\log \left(2 \right)}{2} \right]~-~\left[0 \right]}$

$\small{~=~\frac{\log \left(2 \right)}{2} }$

Solved Example 23.90
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{2}}{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$

$\small{\Rightarrow F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(\cos^2 \phi \right)^2 \right]d \phi}}$

$\small{\Rightarrow F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(1 - \sin^2 \phi \right)^2 \right]d \phi}}$

Put $\small{u~=~\sin \phi \Rightarrow \frac{du}{d\phi}~=~\cos \phi \Rightarrow \cos \phi\,d\phi~=~du}$

So we have: $\small{F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(1 - \sin^2 \phi \right)^2 \right]d\phi}~=~\int{\left[\sqrt{u}\,\left(1 - u^2 \right)^2 \right]du}}$

$\small{~=~\int{\left[\sqrt{u}\,\left(1 - 2 u^2 +  u^4 \right) \right]du}~=~\int{\left[u^{1/2}~-~2 u^{5/2}~+~u^{9/2} \right]du}}$

$\small{~=~\frac{u^{3/2}}{3/2}~-~\frac{2 u^{7/2}}{7/2}~+~\frac{u^{11/2}}{11/2}~=~\frac{2 u^{3/2}}{3}~-~\frac{4 u^{7/2}}{7}~+~\frac{2 u^{11/2}}{11}}$

2. We wrote: $\small{u~=~\sin \phi}$

   ♦ When $\small{\phi}$ approach the lower limit zero, u approach 0

   ♦ When $\small{\phi}$ approach the upper limit $\small{\frac{\pi}{2}}$, u approach 1

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{1}{\left[u^{1/2}~-~2 u^{5/2}~+~u^{9/2} \right]d \phi}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[\frac{2 u^{3/2}}{3}~-~\frac{4 u^{7/2}}{7}~+~\frac{2 u^{11/2}}{11} \right]_{0}^{1}}$

$\small{~=~\left[\frac{2 }{3}~-~\frac{4}{7}~+~\frac{2 }{11} \right]~-~\left[0 \right]}$

$\small{~=~\frac{64}{231} }$

Solved Example 23.91
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$

Put $\small{u~=~\tan^{-1}x \Rightarrow \frac{du}{dx}~=~\frac{1}{1+x^2} \Rightarrow \frac{dx}{1+x^2}~=~du}$

So we have: $\small{F~=~\int{\left[u \right]du}~=~\frac{u^2}{2}}$

2. We wrote: $\small{u~=~\tan^{-1}x}$

   ♦ When $\small{x}$ approach the lower limit zero, u approach 0

   ♦ When $\small{x}$ approach the upper limit $\small{1}$, u approach $\small{\frac{\pi}{4}}$

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[u \right]du}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\frac{u^2}{2} \right]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{\pi^2 }{32} \right]~-~\left[0 \right]~=~\frac{\pi^2 }{32}}$

Solved Example 23.92
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$  

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$

Put $\small{x~=~\tan u}$

$\small{\Rightarrow \sin^{-1}\left(\frac{2x}{1+x^2} \right)~=~\sin^{-1}\left(\frac{2\tan u}{1+\tan^2 u} \right)~=~\sin^{-1}\left(\frac{2\tan u}{\sec^2 u} \right)}$

$\small{~=~\sin^{-1}\left(2 \sin u \cos u \right)~=~\sin^{-1}\left(\sin 2u \right)~=~2u ~=~2 \tan^{-1}x}$

So we have:

$\small{F~=~\int{\left[2 \tan^{-1}x \right]dx}~=~2 \int{\left[\tan^{-1}x \right]dx}~=~2\left(x \tan^{-1}x~+~\frac{\log \left|1+x^2 \right|}{2} \right)}$

(See solved example 23.54 of section 23.17)

$\small{~=~2x \tan^{-1}x~+~\log \left|1+x^2 \right|}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[2x \tan^{-1}x~+~\log \left|1+x^2 \right| \right]_{0}^{1}}$

$\small{~=~\left[2(1) \tan^{-1}1~+~\log \left|1+1^2 \right| \right]~-~\left[2(0) \tan^{-1}(0)~+~\log \left|1+(0)^2 \right| \right]~=~\frac{\pi^2 }{32}}$

$\small{~=~\left[(2)\frac{\pi}{4}~+~\log \left|2 \right| \right]~-~\left[0 \right]~=~\frac{\pi}{2}~+~\log (2) }$

Solved Example 23.93
Evaluate the definite integral $\small{\int_{0}^{2}{\left[x \sqrt{x+2} \right]dx}}$ (Put x+2 =  t²)
Solution:
1. Let $\small{I~=~\int_{0}^{2}{\left[x \sqrt{x+2} \right]dx}}$

Put $\small{x+2} =  t^2\Rightarrow x= t^2 - 2\Rightarrow\frac{dx}{dt}~=~2t \Rightarrow dx = 2t\, dt$

   ♦ When x approaches zero, t approaches $\small{\sqrt 2}$

   ♦ When x approaches 2, t approaches 2

So we get: $\small{F ~=~ \int_{\sqrt 2}^{2}{\left[(t^2 - 2) \sqrt{t^2} \right]2t\,dt}~=~ \int_{\sqrt 2}^{2}{\left[(t^2 - 2) t \right]2t\,dt}}$

$\small{~=~ \int_{\sqrt 2}^{2}{\left[2t^4 - 4t^2 \right]dx}~=~\frac{2 t^5}{5}~-~\frac{4t^3}{3}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(\sqrt 2)~=~\left[\frac{2 t^5}{5}~-~\frac{4t^3}{3} \right]_{\sqrt 2}^{2}}$

$\small{~=~\left[\frac{2 (2)^5}{5}~-~\frac{4(2)^3}{3} \right]~-~\left[\frac{2 (\sqrt 2)^5}{5}~-~\frac{4(\sqrt 2)^3}{3} \right]}$

$\small{~=~\left[\frac{2^6}{5}~-~\frac{2^5}{3} \right]~-~\left[\frac{2^3 (\sqrt 2)}{5}~-~\frac{2^3(\sqrt 2)}{3} \right]}$

$\small{~=~\frac{2^6}{5}~-~\frac{2^5}{3} ~-~\frac{2^3 (\sqrt 2)}{5}~+~\frac{2^3(\sqrt 2)}{3} }$

$\small{~=~\frac{(3)2^6~-~(5)2^5~-~(3)2^3(\sqrt 2)~+~(5)2^3(\sqrt 2)}{15}}$

$\small{~=~\frac{(3)2^6~-~(5)2^5~+~(2)2^3(\sqrt 2)}{15}~=~\frac{(3)2^6~-~(5)2^5~+~2^4(\sqrt 2)}{15}}$

$\small{~=~\frac{2^5 \left[6~-~5 \right]~+~2^4(\sqrt 2)}{15}~=~\frac{2^5 ~+~2^4(\sqrt 2)}{15}~=~\frac{2^4 \left[2 ~+~\sqrt 2 \right]}{15}}$

$\small{~=~\frac{16 \sqrt 2 \left[\sqrt 2 ~+~1 \right]}{15}}$

Solved Example 23.94
Evaluate the definite integral $\small{\int_{0}^{2}{\left[\frac{1}{x+4 - x^2} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{2}{\left[\frac{1}{x+4 - x^2} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{x+4 - x^2} \right]dx}}$

• In our present case, a = -1, b = 1 and c = 4

2. So we can calculate u and $\small{k^2}$:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,\frac{1}{2(-1)}~=~(x-\frac{1}{2})}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{4}{-1}\,-\,\frac{(1)^2}{4(-1)^2}~=~-4\,-\,\frac{1}{4}~=~\frac{-17}{4}}$

3. So we want:

$\small{\int{\left[\frac{dx}{x+4 - x^2} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{-1}\int{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}~=~(-1)\int{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. That means, we want the definite integral:

$\small{~(-1)\int_0^2{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}}$

(i) Put t = (x−1/2). Then dt/dx = 1, which gives dt = dx

   ♦ Also, when x approaches zero, t approaches -(1/2)

   ♦ Similarly, when x approaches 2, t approaches (3/2)

• So we want:

$\small{~(-1)\int_0^2{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}~=~~(-1)\int_{-1/2}^{3/2}{\left[\frac{dt}{(t)^2~-~\left(\frac{\sqrt{17}}{2} \right)^2} \right]}}$

(ii) We have formula: $\small{\int{\left[\frac{dt}{t^2\,-\,m^2} \right]}\,=\, \frac{1}{2m} \log \left| \frac{t-m}{t+m}  \right|}$

• In our present case, m = $\small{\frac{\sqrt{17}}{2}}$

5. So we get:
$\small{(-1)\int{\left[\frac{dt}{(t)^2~-~\left(\frac{\sqrt{17}}{2} \right)^2}  \right]}\,=\, \frac{-1}{\sqrt{17}} \log \left| \frac{t-\frac{\sqrt{17}}{2}}{t+\frac{\sqrt{17}}{2}}  \right|}$

6. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(-1/2)\,-\,F(3/2)~=~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{t-\frac{\sqrt{17}}{2}}{t+\frac{\sqrt{17}}{2}}  \right| \right]_{-1/2}^{3/2}}$

$\small{~=~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{\frac{3}{2}-\frac{\sqrt{17}}{2}}{\frac{3}{2}+\frac{\sqrt{17}}{2}}  \right| \right]~-~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{\frac{-1}{2}-\frac{\sqrt{17}}{2}}{\frac{-1}{2}+\frac{\sqrt{17}}{2}}  \right|\right]}$

$\small{~=~\left[\frac{-1}{\sqrt{17}} \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]~+~\left[\frac{1}{\sqrt{17}} \log \left|\frac{-1 - \sqrt{17}}{-1 + \sqrt{17}}  \right|\right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{-1 - \sqrt{17}}{-1 + \sqrt{17}}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{(-1)(1 + \sqrt{17})}{(-1)(1 - \sqrt{17})}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{1 + \sqrt{17}}{1 - \sqrt{17}}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\big[\log\left[(1 + \sqrt{17})(3 + \sqrt{17}) \right]~-~\log\left[(1 - \sqrt{17})(3 - \sqrt{17}) \right] \big]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\big[\log\left[20 + 4 \sqrt {17} \right]~-~\log\left[20 - 4 \sqrt{17} \right] \big]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{4(5 + \sqrt{17})}{4(5 - \sqrt{17})} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{5 + \sqrt{17}}{5 - \sqrt{17}} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{(5 + \sqrt{17})(5 + \sqrt{17})}{(5 - \sqrt{17})(5 + \sqrt{17})} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{25 + 10 \sqrt{17} + 17}{25 - 17} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{42 + 10 \sqrt{17}}{8} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{21 + 5 \sqrt{17}}{4} \right]  \bigg]}$

Solved Example 23.95
Evaluate the definite integral $\small{\int_{-1}^{1}{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^{1}{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$

• In our present case, a = 1, b = 2 and c = 5

2. So we can calculate u and $\small{k^2}$:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,1}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{5}{1}\,-\,\frac{(2)^2}{4(1)^2}~=~5\,-\,1~=~4}$

3. So we want:

$\small{\int{\left[\frac{dx}{x^2 + 2x + 5} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{1}\int{\left[\frac{dx}{(x+1)^2~+~2^2} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. That means, we want the definite integral:

$\small{\int_{-1}^1{\left[\frac{dx}{(x+1)^2~+~2^2} \right]}}$

(i) Put t = (x+1). Then dt/dx = 1, which gives dt = dx

   ♦ Also, when x approaches -1, t approaches zero

   ♦ Similarly, when x approaches 1, t approaches 2

• So we want:

$\small{\int_{-1}^1{\left[\frac{dx}{(x+1)^2~+~2^2} \right]}~=~~\int_{0}^{2}{\left[\frac{dt}{(t)^2~+~\left(2 \right)^2} \right]}}$

(ii) We have formula: $\small{\int{\left[\frac{dt}{t^2\,+\,m^2} \right]}\,=\, \frac{1}{m} \tan^{-1}\left( \frac{t}{m} \right)}$

• In our present case, m = 2

5. So we get:
$\small{\int{\left[\frac{dt}{(t)^2~+~\left(2 \right)^2}  \right]}\,=\, \frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right)}$

6. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right) \right]_{0}^{2}}$

$\small{~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{2}{2} \right) \right]~-~\left[\frac{1}{2} \tan^{-1}\left( \frac{0}{2} \right)\right]}$

$\small{~=~\left[\frac{1}{2} \tan^{-1}\left( 1 \right) \right]~-~\left[\frac{1}{2} \tan^{-1}\left(0 \right)\right]}$

$\small{~=~\left[\frac{1}{2} \left(\frac{\pi}{4} \right) \right]~-~\left[\frac{1}{2} \left(0 \right)\right]}$

$\small{~=~\frac{\pi}{8}}$

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The link below gives a few more solved examples:

Exercise 23.10

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In the next section, we will see a few more solved examples.

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23.24 - Solved Examples on Evaluation of Definite Integrals

In the previous section, we saw the fundamental theorem of calculus. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.82
Evaluate the definite integral $\small{\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\left[\csc(x) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\left[\csc(x) \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\csc(x) \right]dx}~=~\log \left| \csc (x)~-~\cot (x)\right|}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{6})\,-\,F(\frac{\pi}{4})~=~\left[\log \left| \csc (x)~-~\cot (x)\right| \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}}$

$\small{~=~\left[\log \left| \csc (\frac{\pi}{4})~-~\cot (\frac{\pi}{4})\right| \right]~-~\left[\log \left| \csc (\frac{\pi}{6})~-~\cot (\frac{\pi}{6})\right| \right]}$

$\small{~=~\left[\log \left|\sqrt{2}-1 \right| \right]~-~\left[\log \left|2 - \sqrt 3 \right| \right]~=~\log \left|\frac{\sqrt 2 - 1}{2 - \sqrt 3} \right|}$

3. The required definite integral is equal to the area of the blue region in fig.23.20 below:

Fig.23.20

Solved Example 23.83
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{1}{\sqrt{1-x^2}} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{1}{\sqrt{1-x^2}} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{\sqrt{1-x^2}} \right]dx}~=~\sin^{-1}x}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[\sin^{-1}x \right]_{0}^{1}}$

$\small{~=~\left[\sin^{-1}(1) \right]~-~\left[\sin^{-1}(0) \right]~=~\frac{\pi}{2}~-~0}$

$\small{~=~\frac{\pi}{2}}$

3. The required definite integral is equal to the area of the blue region in fig.23.21 below:

Fig.23.21

Solved Example 23.84
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{1}{1+x^2} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{1}{1+x^2} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{1+x^2} \right]dx}~=~\tan^{-1}x}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[\tan^{-1}x \right]_{0}^{1}}$

$\small{~=~\left[\tan^{-1}(1) \right]~-~\left[\tan^{-1}(0) \right]~=~\frac{\pi}{4}~-~0}$

$\small{~=~\frac{\pi}{4}}$

3. The required definite integral is equal to the area of the blue region in fig.23.22 below:

Fig.23.22

Solved Example 23.85
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{1}{x^2 - 1} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{2}^{3}{\left[\frac{1}{x^2 - 1} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{\int{\left[\frac{1}{x^2 - a^2} \right]dx}~=~\frac{1}{2a} \log \left| \frac{x-a}{x+a}  \right|}$

So we get:
$\small{F~=~\int{\left[\frac{1}{x^2 - 1} \right]dx}~=~\frac{1}{2(1)} \log \left| \frac{x-1}{x+1}  \right|}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(3)\,-\,F(2)~=~\left[\frac{1}{2} \log \left| \frac{x-1}{x+1}  \right| \right]_{2}^{3}}$

$\small{~=~\left[\frac{1}{2} \log \left| \frac{3-1}{3+1}  \right| \right]~-~\left[\frac{1}{2} \log \left| \frac{2-1}{2+1}  \right| \right]}$

$\small{~=~\left[\frac{1}{2} \log \left| \frac{2}{4}  \right| \right]~-~\left[\frac{1}{2} \log \left| \frac{1}{3}  \right| \right]}$

$\small{~=~\frac{1}{2} \left[ \log \left| \frac{1}{2}  \right|~-~\log \left| \frac{1}{3}  \right| \right]}$

$\small{~=~\frac{1}{2} \left[ \log \left| \frac{1/2}{1/3}  \right| \right]}$

$\small{~=~\frac{1}{2} \left[ \log \left( \frac{3}{2}  \right) \right]}$

3. The required definite integral is equal to the area of the blue region in fig.23.23 below:

Fig.23.23

Solved example 23.86
Evaluate the following integrals:
$\small{\text{(i)}~\int_{2}^{3}{\left[x^2 \right]dx}~~~~~~~\text{(ii)}~\int_{4}^{9}{\left[\frac{\sqrt x}{\left(30 - x^{\frac{3}{2}} \right)^2} \right]dx}}$

$\small{\text{(iii)}~\int_{1}^{2}{\left[\frac{x}{(x+1)(x+2)} \right]dx}~~~~~~~\text{(iv)}~\int_{0}^{\frac{\pi}{4}}{\left[\sin^3(2t) \cos(2t) \right]dt}}$
Solution:
Part (i): $\small{\int_{2}^{3}{\left[x^2 \right]dx}}$

1. Let $\small{I~=~\int_{2}^{3}{\left[x^2 \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{\int{\left[x^2 \right]dx}~=~\frac{x^3}{3}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(3)\,-\,F(2)~=~\left[\frac{x^3}{3} \right]_{2}^{3}}$

$\small{~=~\left[\frac{3^3}{3}\right]~-~\left[\frac{2^3}{3} \right]}$

$\small{~=~\left[\frac{27}{3} \right]~-~\left[\frac{8}{3} \right]~=~\frac{19}{3}}$

Part (ii): $\small{\int_{4}^{9}{\left[\frac{\sqrt x}{\left(30 - x^{\frac{3}{2}} \right)^2} \right]dx}}$

1. Let $\small{I~=~\int_{4}^{9}{\left[\frac{\sqrt x}{\left(30 - x^{\frac{3}{2}} \right)^2} \right]dx}}$

• First we find the indefinite integral F.

Let $\small{u = 30 - x^{\frac{3}{2}}}$. Then $\small{\frac{du}{dx}~=~(-1)\frac{3}{2} x^{1/2}}$

$\small{\Rightarrow du ~=~(-1)\frac{3}{2} x^{1/2}\,dx}$

$\small{F~=~\int{\left[\frac{(-1)(-1) (3/2)(2/3) x^{1/2}}{\left(30 - x^{\frac{3}{2}} \right)^2} \right]dx}~=~\int{\left[\frac{(-1)(2/3)du}{\left(u \right)^2} \right]}}$

$\small{~=~(-1)(2/3)\,\int{\left[\frac{du}{\left(u \right)^2} \right]}~=~(-1)(2/3){\left[\frac{u^{-1}}{\left(-1 \right)} \right]}}$

$\small{~=~\frac{2}{3u}~=~\frac{2}{3\left(30 - x^{\frac{3}{2}}\right)}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(9)\,-\,F(4)~=~\left[\frac{2}{3\left(30 - x^{\frac{3}{2}}\right)} \right]_{4}^{9}}$

$\small{~=~\left[\frac{2}{3\left(30 - 9^{\frac{3}{2}}\right)}\right]~-~\left[\frac{2}{3\left(30 - 4^{\frac{3}{2}}\right)} \right]}$

$\small{~=~\left[\frac{2}{3\left(3\right)}\right]~-~\left[\frac{2}{3\left(22\right)} \right]~=~\frac{19}{99}}$

Part (iii): $\small{\int_{1}^{2}{\left[\frac{x}{(x+1)(x+2)} \right]dx}}$

1. Let $\small{I~=~\int_{1}^{2}{\left[\frac{x}{(x+1)(x+2)} \right]dx}}$

• First we find the indefinite integral F.

(a) The numerator is a polynomial of degree one. The denominator is a polynomial of degree 2.

(b) So it is a proper rational function. We can straight away start partial fraction decomposition

(c) The denominator is already in the factorized form:
(x+1)(x+2)

   ♦ All factors are linear
   
   ♦ And all factors are distinct from one another.

(d) So we are able to write:

$\small{\frac{x}{(x+1)(x+2)}\,=\,\frac{A_1}{x+1}~+~\frac{A_2}{x+2}}$

Where A₁ and A₂ are real numbers.

(e) To find A₁ and A₂, we make denominators same on both sides:

$\small{\frac{x}{(x+1)(x+2)}\,=\,\frac{A_1}{x+1}~+~\frac{A_2}{x+2}~=~\frac{A_1 (x+2)~+~A_2 (x+1)}{(x+1)(x+2)}}$

(f) Since denominators are same on both sides, we can equate the numerators. We get:
$\small{x~=~A_1 (x+2)~+~A_2 (x+1)}$

(g) After equating the numerators, we can use suitable substitution.

   ♦ Put x = −2. We get: −2 = A₂(−1). So A₂ = 2  

♦ Put x = −1. We get: −1 = A₁(1). So A₁ = −1

(h) Now the result in (d) becomes:

$\small{\frac{x}{(x+1)(x+2)}\,=\,\frac{-1}{x+1}~+~\frac{2}{x+2}}$

(h) So the integration becomes easy. We get:

$\small{-\log \left|x+1  \right|~+~2\log \left|x+2  \right|}$


2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(1)~=~\left[-\log \left|x+1  \right|~+~2\log \left|x+2  \right| \right]_{1}^{2}}$

$\small{~=~\left[-\log \left|2+1  \right|~+~2\log \left|2+2  \right|\right]~-~\left[-\log \left|1+1  \right|~+~2\log \left|1+2  \right|\right]}$

$\small{~=~\left[-\log \left|3  \right|~+~2\log \left|4  \right|\right]~-~\left[-\log \left|2  \right|~+~2\log \left|3  \right|\right]}$

$\small{~=~-\log \left|3  \right|~+~2\log \left|4  \right|~+~\log \left|2  \right|~-~2\log \left|3  \right|}$

$\small{~=~-3\log \left|3  \right|~+~4\log \left|2  \right|~+~\log \left|2  \right|}$

$\small{~=~-3\log \left|3  \right|~+~5\log \left|2  \right|~=~\log\left(\frac{32}{27} \right)}$

Part (iv): $\small{\int_{0}^{\frac{\pi}{4}}{\left[\sin^3(2t) \cos(2t) \right]dt}}$

1. Let $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\sin^3(2t) \cos(2t) \right]dt}}$

• First we find the indefinite integral F.

Let $\small{u = \sin(2t)}$. Then $\small{\frac{du}{dx}~=~2 \cos (2t)}$

$\small{\Rightarrow du ~=~2 \cos (2t)\,dx}$

$\small{F~=~\int{\left[(2/2)\sin^3(2t) \cos(2t) \right]dx}~=~\int{\left[(1/2)u^3 \right]du}}$

$\small{~=~(1/2)\,\int{\left[u^3 \right]du}~=~(1/2){\left[\frac{u^{4}}{4} \right]}}$

$\small{~=~\frac{\sin^4(2t)}{8}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\frac{\sin^4(2t)}{8} \right]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{\sin^4(\frac{\pi}{2})}{8}\right]~-~\left[\frac{\sin^4(0)}{8} \right]}$

$\small{~=~\left[\frac{1^4}{8}\right]~-~\left[\frac{0}{8} \right]~=~\frac{1}{8}}$

Solved Example 23.87
Evaluate the definite integral $\small{\int_{1}^{2}{\left[\frac{5x^2}{x^2 + 4x + 3} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{1}^{2}{\left[\frac{5x^2}{x^2 + 4x + 3} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{5x^2}{x^2 + 4x + 3} \right]dx}}$

(a) The numerator is a polynomial of degree 2. The denominator is also a polynomial of degree 2.

(b) So it is not a proper rational function. We must do long division. We get:
$\small{\frac{5x^2}{x^2 + 4x + 3}\,=\,5~-~ \frac{20x + 15}{x^2 + 4x + 3}}$

• The reader may write all steps involved in the long division (or any other suitable method) process.

(c) In the R.H.S, the first term can be easily integrated. But the second term must be subjected to partial fraction decomposition.

• First we factorize the denominator. We get:

$\small{x^2 + 4x + 3~=~(x+1)(x+3)}$

• The reader may write all steps involved in the factorization process.   

   ♦ All factors are linear

   ♦ And all factors are distinct from one another.

(d) So we are able to write:

$\small{\frac{20x + 15}{x^2 + 4x + 3}\,=\,\frac{20x + 15}{(x+1)(x+3)}\,=\,\frac{A_1}{x+1}~+~\frac{A_2}{x+3}}$

Where A₁ and A₂ are real numbers.

(e) To find A₁ and A₂, we make denominators same on both sides:

$\small{\frac{20x + 15}{(x+1)(x+3)}\,=\,\frac{A_1}{x+1}~+~\frac{A_2}{x+3}~=~\frac{A_1 (x+3)~+~A_2 (x+1)}{(x+1)(x+3)}}$

(f) Since denominators are same on both sides, we can equate the numerators. We get:
$\small{20x + 15~=~A_1 (x+3)~+~A_2 (x+1)}$

(g) After equating the numerators, we can use suitable substitution.
   
   ♦ Put x = -1. We get: -5 = A₁(2). So A₁ = −5/2
   
   ♦ Put x = -3. We get: -45 = A₂(-2). So A₂ = 45/2

(h) Now the result in (b) becomes:

$\small{\frac{5x^2}{x^2 + 4x + 3}\,=\,5~-~ \frac{20x + 15}{x^2 + 4x + 3}~=~5~-~\left[\frac{(-5)}{2(x+1)}~+~\frac{45}{2(x+3)} \right]}$

$\small{~=~5~+~\frac{5}{2(x+1)}~-~\frac{45}{2(x+3)} }$

(i) So the integration becomes easy. We get:
$\small{5x + \frac{5}{2} \log \left|x+1 \right| -\frac{45}{2} \log \left|x+3 \right|}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{6})\,-\,F(\frac{\pi}{4})~=~\left[5x + \frac{5}{2} \log \left|x+1 \right| -\frac{45}{2} \log \left|x+3 \right| \right]_{1}^{2}}$

$\small{~=~\left[10 + \frac{5}{2} \log \left|3 \right| -\frac{45}{2} \log \left|5 \right|\right]~-~\left[5 + \frac{5}{2} \log \left|2 \right| -\frac{45}{2} \log \left|4 \right| \right]}$

$\small{~=~10 + \frac{5}{2} \log \left|3 \right| -\frac{45}{2} \log \left|5 \right|~-~5 - \frac{5}{2} \log \left|2 \right| +\frac{45}{2} \log \left|4 \right| }$

$\small{~=~5 + \frac{5}{2} \log \left|\frac{3}{2} \right|  +\frac{45}{2} \log \left|\frac{4}{5} \right| }$

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The link below gives a few more solved examples:

Exercise 23.9

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In the next section, we will see evaluation of definite integrals by substitution.

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23.23 - Fundamental Theorem of Calculus

In the previous section, we completed a discussion on definite integral as limit of a sum. In this section, we will see an easier method to calculate the definite integral.

Some basics can be written in 4 steps:
1. We have seen that, $\small{\int_a^b{\left[f(x) \right]dx}}$ is equal to the area bounded by four items:
   ♦ The curve y = f(x)
   ♦ The vertical line x = a
   ♦ The vertical line x = b
   ♦ The horizontal line y = 0 (x-axis)
• This area is marked by rightward strokes in the fig.23.10 below:

Fig.23.10

2. Consider the interval [a,b] on the x-axis. Take any arbitrary point x from within the interval [a,b].
• We can draw a vertical line through x. Then we get an area bounded by the four items:
   ♦ The curve y = f(x)
   ♦ The vertical line x = a
   ♦ The vertical line through the arbitrary point x
   ♦ The horizontal line y = 0 (x-axis)
• This area is marked by both rightward strokes and leftward strokes in the fig.23.10 above.
• Obviously, this area is equal to $\small{\int_a^x{\left[f(x) \right]dx}}$
3. The area mentioned in step (3) above depends on the value of x. So this area is a function of x.
• We denote this function of x as A(x).
• A(x) is called the area function and is given by:
$\small{A(x)~=~\int_a^x{\left[f(x) \right]dx}}$
4. In fig.23.10 above, it is assumed that f(x) is +ve in the interval [a,b].
• However, the above four steps are valid for all functions in [a,b]

• For example, in fig.23.11 below, consider the area bounded by the four items:
   ♦ The curve y = f(x) =0.1x³ −1.6
   ♦ The vertical line x = −2
   ♦ The vertical line x = 3
   ♦ The horizontal line y = 0 (x-axis)
• This area is filled in blue color.

Fig.23.11

• The area of the blue region will be equal to:

$\small{\int_{-2}^3 {\left[0.1x^3~-~1.6 \right]dx}}$

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First fundamental theorem of integral calculus

• Let f be a continuous function in [a,b] and let A(x) be the area function.
• Then the first theorem states that, the integral of f will be equal to the derivative of the area function.
That is.,$\small{A'(x)~=~f(x)}$ for all x ∈ [a,b]
We will see the proof in higher classes.

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Second fundamental theorem of integral calculus

• Let f be a continuous function defined in [a,b] and F be the anti derivative of f.
• Then the second theorem states that the definite integral $\small{\int_a^b{\left[f(x) \right]dx}}$ will be equal to $\small{F(b)\,-\,F(a)}$.
• $\small{F(b)\,-\,F(a)}$ is denoted as $\small{[F(x)]_a^b}$
• So we can write:
$\small{\int_a^b{\left[f(x) \right]dx}~=~[F(x)]_a^b~=~F(b)\,-\,F(a)}$
We will see the proof in higher classes.

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Two important points about the second theorem is written below:
1.  To determine $\small{\int_a^b{\left[f(x) \right]dx}}$, first we find $\small{F(x)+C}$, which is the indefinite integral of f(x)
• Then we calculate $\small{[F(x) + C]_a^b}$ which is given by $\small{[F(b)\, + C]-\,[F(a) + C]}$
• That is., $\small{\int_a^b{\left[f(x) \right]dx}~=~[F(x) + C]_a^b~=~F(b)\,-\,F(a)}$
• Note that, the constant of integration C gets cancelled. So there is no need to write C in the calculation steps.

2. Consider the definite integral $\small{\int_{-2}^3 \left[{x \sqrt{x^2 - 1}}\right]dx}$
• Here we are dealing with every points in the interval [−2,3]
• (−1,1) is an interval which lies within [−2,3].
• Take any point from (−1,1). At that point, f(x) is not defined because, it involves the square root of a −ve number. The graph is shown in fig.23.12 below:

Fig.23.12

• That means, f(x) is not well defined in [−2,3]. So it is erraneous to consider $\small{\int_{-2}^3{\left[x \sqrt{x^2 - 1} \right]dx}}$
• It is clear that, to find the definite integral of a function f in the interval [a,b], that f must be well defined and continuous in [a,b]

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Let us see a solved example:
Solved Example 23.75
Evaluate the definite integral $\small{\int_{-1}^1{\left[x+1 \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^1{\left[x+1 \right]dx}}$
• First we find the indefinite integral F. We have:
$\small{F~=~\int{\left[x+1 \right]dx}~=~\frac{x^2}{2}\,+\,x}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(-1)~=~\left[\frac{x^2}{2}\,+\,x \right]_{-1}^1}$

$\small{~=~\left[\frac{1^2}{2}\,+\,1 \right]~-~\left[\frac{(-1)^2}{2}\,+\,(-1) \right]}$

$\small{~=~\left[\frac{3}{2} \right]~-~\left[\frac{-1}{2} \right]~=~2}$

3. A check can be done in 3 steps:
(i) In fig.23.13 below, the triangle is filled with blue color.

Fig.23.13

• The area of the triangle will be equal to the required definite integral.
(ii) The triangle has a base of 2 units.
    ♦ The red line has the equation y = x+1
    ♦ The right side green line has the equation x = 1
• Solving the two equations, we get:x = 1, y = 2
• So height of the triangle is 2 units.
(iii) Then the area of the triangle = $\small{\frac{1}{2}(2)(2)~=~2~ \text{square units}}$

Solved Example 23.76
Evaluate the definite integral $\small{\int_{2}^3{\left[\frac{1}{x} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{2}^3{\left[\frac{1}{x} \right]dx}}$
• First we find the indefinite integral F. We have:
$\small{F~=~\int{\left[\frac{1}{x} \right]dx}~=~\log x}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(3)\,-\,F(2)~=~\left[\log x \right]_{2}^3}$

$\small{~=~\left[\log 3 \right]~-~\left[\log 2 \right]}$

$\small{~=~\log \left(\frac{3}{2} \right)}$

3. The required definite integral is equal to the area of the blue region in fig.23.14 below:

Fig.23.14

Solved Example 23.77
Evaluate the definite integral $\small{\int_{1}^2{\left[4x^3 - 5x^2 + 6x + 9 \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{1}^2{\left[4x^3 - 5x^2 + 6x + 9 \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[4x^3 - 5x^2 + 6x + 9 \right]dx}}$

$\small{~=~x^4 - \frac{5x^3}{3} + 3x^2 + 9x}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(1)~=~\left[x^4 - \frac{5x^3}{3} + 3x^2 + 9x \right]_{1}^2}$

$\small{~=~\left[2^4 - \frac{(5)2^3}{3} + (3)2^2 + (9)2 \right]~-~\left[1^4 - \frac{(5)1^3}{3} + (3)1^2 + (9)1 \right]}$

$\small{~=~\left[16 - \frac{40}{3} + 12 + 18 \right]~-~\left[1 - \frac{5}{3} + 3 + 9 \right]}$

$\small{~=~\left[\frac{98}{3} \right]~-~\left[\frac{34}{3} \right]}$

$\small{~=~\frac{64}{3}}$

3. The required definite integral is equal to the area of the blue region in fig.23.15 below:

Fig.23.15

Solved Example 23.78
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{4}}{\left[\sin(2x) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\sin(2x) \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\sin(2x) \right]dx}~=~\frac{-\cos(2x)}{2}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\frac{-\cos(2x)}{2} \right]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{-\cos(\frac{\pi}{2})}{2} \right]~-~\left[\frac{-\cos(0)}{2} \right]}$

$\small{~=~\left[0 \right]~+~\left[\frac{1}{2} \right]}$

$\small{~=~\frac{1}{2}}$

3. The required definite integral is equal to the area of the blue region in fig.23.16 below:

Fig.23.16

Solved Example 23.79
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{2}}{\left[\cos(2x) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\cos(2x) \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\cos(2x) \right]dx}~=~\frac{\sin(2x)}{2}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{2})\,-\,F(0)~=~\left[\frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}}}$

$\small{~=~\left[\frac{\sin(\pi)}{2} \right]~-~\left[\frac{\sin(0)}{2} \right]}$

$\small{~=~\left[0 \right]~-~\left[0 \right]}$

$\small{~=~0}$

3. The required definite integral is equal to the area of the blue region in fig.23.17 below:

Fig.23.17

The upper blue region and lower blue region, have the same area. But they are opposite in sign. So the net area is zero. 

Solved Example 23.80
Evaluate the definite integral $\small{\int_{4}^{5}{\left[e^x \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{4}^{5}{\left[e^x \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[e^x \right]dx}~=~e^x}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(5)\,-\,F(4)~=~\left[e^x \right]_{4}^{5}}$

$\small{~=~\left[e^5 \right]~-~\left[e^4 \right]}$

$\small{~=~e^4(e-1)}$

3. The required definite integral is equal to the area of the blue region in fig.23.18 below:

Fig.23.18

Solved Example 23.81
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{4}}{\left[\tan(x) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\tan(x) \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\tan(x) \right]dx}~=~\log \left| \sec x\right|}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\log \left| \sec x\right| \right]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left[\log \left| \sec (\frac{\pi}{4})\right| \right]~-~\left[\log \left| \sec (0)\right| \right]}$

$\small{~=~\left[\log \left|\sqrt{2} \right| \right]~-~\left[\log \left|1 \right| \right]~=~\log \left|\frac{\sqrt 2}{1} \right|~=~\log \left|\sqrt 2 \right|}$

$\small{~=~\frac{1}{2} \log \left|2 \right|}$

3. The required definite integral is equal to the area of the blue region in fig.23.19 below:

Fig.23.19

---

In the next section, we will see a few more solved examples.

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23.22 - Solved Examples on Definite Integral as Limit of Sum

In the previous section, we saw the definite integral as limit of sum. We saw a solved example also. In this section, we will see a few more solved examples.

Solved Example 23.70
Evaluate $\small{\int_0^5{\left[x+1 \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^5{\left[x+1 \right]dx}}$ is the area bounded by the four items:
   ♦ The line y = f(x) = x+1.
   ♦ The vertical line x = 0
   ♦ The vertical line x = 5
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.5 below. Our task is to find the area of the violet shaded area.

Fig.23.5

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 5
So $\small{h~=~\frac{b-a}{n}~=~\frac{5-0}{n}~=~\frac{5}{n}}$

3. Now the formula becomes:
$\small{\int_0^5{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{5-0}{n} \left[f(a+\frac{5}{n})~+~f(a+\frac{10}{n})~+~~.~.~.~+f(a+\frac{5n}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{5}{n} \left[(a+\frac{5}{n}\,+1)~+~(a+\frac{10}{n}\,+1)~+~.~.~.~+(a+\frac{5n}{n}+1) \right]}$

4. Let us determine the quantity inside the square brackets:

$\small{ \left[(a+\frac{5}{n}\,+1)~+~(a+\frac{10}{n}\,+1)~+~.~.~.~+(a+\frac{5n}{n}+1) \right]}$

5. So we have to do three summations:

(i) $\small{a~+~a~+~a~+~.~.~.~ \text{n terms}~=~na~=~n(0)~=~0}$

(ii) $\small{\frac{5}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{5}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{5(n+1)}{2}~=~\frac{5n}{2}~+~\frac{5}{2}}$

(Here we use the technique of arithmetic progression)

(iii) $\small{1~+~1~+~1~+~.~.~.~ \text{n terms}~=~n(1)~=~n}$

6. So the total of three summations is:

$\small{0~+~\frac{5n}{2}~+~\frac{5}{2}~+~n}$

7. Now the limit in step (3) can be calculated:

$\small{\lim_{n\rightarrow \infty} \frac{5}{n} \left[\frac{5n}{2}~+~\frac{5}{2}~+~n \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \left[\frac{25}{2}~+~\frac{25}{2n}~+~5 \right]~=~\frac{35}{2}}$

Solved Example 23.71
Evaluate the definite integral $\small{\int_0^2{\left[x^2+1 \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^2{\left[x^2+1 \right]dx}}$ is the area bounded by the four items:
   ♦ The line y = f(x) = x²+1.
   ♦ The vertical line x = 0
   ♦ The vertical line x = 2
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.6 below. Our task is to find the area of the violet shaded area.

Fig.23.6

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 2
So $\small{h~=~\frac{b-a}{n}~=~\frac{2-0}{n}~=~\frac{2}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2-0}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \bigg[\big[(a+\frac{2}{n})^2~+~1\big]~+~\big[(a+\frac{4}{n})^2~+~1 \big]~+~\big[(a+\frac{6}{n})^2~+~1 \big]~+~~.~.~.}$

$\small{~~~~~~~~~~.~.~.~+~\big[(a+\frac{2n}{n})^2~+~1 \big] \bigg]}$

4. Let us determine the quantity inside the large square brackets:

$\small{~=~ \bigg[\big[a^2 + \frac{4a}{n}+\frac{2^2}{n^2}~+~1\big]~+~\big[a^2 + \frac{8a}{n}+\frac{4^2}{n^2}~+~1 \big]~+~\big[a^2 + \frac{12a}{n}+\frac{6^2}{n^2}~+~1 \big]~+~ }$

$\small{~~~~~~~~.~.~.~+~\big[a^2 + \frac{2(2an)}{n}+\frac{2^2\,n^2}{n^2}~+~1 \big] \bigg]}$

5. So we have to do four summations:

(i) $\small{a^2~+~a^2~+~a^2~+~.~.~.~ \text{n terms}~=~\small{n a^2~=~n(0^2)~=~0}}$

(ii) $\small{\frac{4a}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)~~~~~=~0, ~~~\because a~=~0}$

(iii) $\small{\frac{2^2}{n^2}\left(1^2~+~2^2~+~3^2~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{2^2}{n^2}\left(\frac{2n^3 + 3 n^2 + n}{6} \right)~=~\frac{4n}{3}~+~2~+~\frac{2}{3n}}$

(Here we use the technique that we saw in section 9.5)

(iv) $\small{1~+~1~+~1~+~.~.~.~ \text{n terms}~=~n (1)~=~n}$

6. So the total of 4 summations is:

$\small{0+0+\frac{4n}{3}~+~2~+~\frac{2}{3n}~+~n}$

8. Now the limit in step (3) can be calculated:

$\small{\lim_{n\rightarrow \infty} \frac{2}{n} \left[\frac{4n}{3}~+~2~+~\frac{2}{3n}~+~n \right]}$

$\small{~=~\lim_{n\rightarrow \infty}  \left[\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}+2 \right]~=~\frac{8}{3}+2~=~\frac{14}{3}}$

Solved Example 23.72
Evaluate the definite integral $\small{\int_0^2{\left[e^x \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^2{\left[e^x \right]dx}}$ is the area bounded by the four items:
   ♦ The curve y = f(x) = e^x.
   ♦ The vertical line x = 0 (y-axis)
   ♦ The vertical line x = 2
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.7 below. Our task is to find the area of the violet shaded area.

Fig.23.7

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = 0 and b = 2
So $\small{h~=~\frac{b-a}{n}~=~\frac{2-0}{n}~=~\frac{2}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2-0}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.~+~e^{a+\frac{2n}{n}}\big]}$

4. Let us determine the quantity inside the large square brackets:

$\small{\big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.+~e^{a+\frac{2n}{n}}\big]}$

$\small{~=~e^a\big[e^{\frac{2}{n}}~+~e^{\frac{4}{n}}~+~e^{\frac{6}{n}}~+~~.~.~.+~e^{\frac{2n}{n}}\big]}$

Here $\small{e^a~=~1~~\because a = 0}$

5. The n terms inside the square brackets, form a geometric progression.

• First term = $\small{e^{\frac{2}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{4}{n}}}{e^{\frac{2}{n}}}~=~\frac{e^{\frac{6}{n}}}{e^{\frac{4}{n}}}~=~e^{\frac{2}{n}}}$

6. Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^n~-~1 \right)}{r~-~1}}$

• So we get:

Sum of all terms inside the brackets

$\small{~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{\frac{2n}{n}}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}}$

7. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{2}{n} \big[\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty}  \Bigg[\frac{e^{\frac{2}{n}} \left({e^{2}}~-~1 \right)}{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$

$\small{~=~\frac{e^{0} \left({e^{2}}~-~1 \right)}{1} ~=~e^2 ~-~1}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{2}{n}} \right]~=~e^{\frac{2}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{2}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$

Solved Example 23.73
Evaluate the definite integral $\small{\int_{-1}^{1}{\left[e^x \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_{-1}^{1}{\left[e^x \right]dx}}$ is the area bounded by the four items:
   ♦ The curve y = f(x) = e^x.
   ♦ The vertical line x = −1
   ♦ The vertical line x = 1
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.8 below. Our task is to find the area of the violet shaded area.

Fig.23.8

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = -1 and b = -1

So $\small{h~=~\frac{b-a}{n}~=~\frac{1-(-1)}{n}~=~\frac{2}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.~+~e^{a+\frac{2n}{n}}\big]}$

4. Let us determine the quantity inside the large square brackets:

$\small{\big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.+~e^{a+\frac{2n}{n}}\big]}$

$\small{~=~e^a\big[e^{\frac{2}{n}}~+~e^{\frac{4}{n}}~+~e^{\frac{6}{n}}~+~~.~.~.+~e^{\frac{2n}{n}}\big]}$

Here $\small{e^a~=~\frac{1}{e}~~\because a = -1}$

5. The n terms inside the square brackets, form a geometric progression.

• First term = $\small{e^{\frac{2}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{4}{n}}}{e^{\frac{2}{n}}}~=~\frac{e^{\frac{6}{n}}}{e^{\frac{4}{n}}}~=~e^{\frac{2}{n}}}$

6. Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^n~-~1 \right)}{r~-~1}}$

• So we get:

Sum of all terms inside the brackets

$\small{~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{\frac{2n}{n}}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}}$

7. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{2}{n}\left(\frac{1}{e} \right) \big[\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n}\left(\frac{{e^{2}}~-~1}{e} \right) \big[\frac{e^{\frac{2}{n}}}{e^{\frac{2}{n}}~-~1} \big]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)~\lim_{n\rightarrow \infty}  \Bigg[\frac{e^{\frac{2}{n}} }{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)~\lim_{n\rightarrow \infty}  \Bigg[\frac{e^{\frac{2}{n}} }{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)  \Big[\frac{1}{1} \Big]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)  ~=~e~-~\frac{1}{e}}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{2}{n}} \right]~=~e^{\frac{2}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{2}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$

Solved Example 23.74
Evaluate the definite integral $\small{\int_{0}^{4}{\left[x+e^{2x} \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_{0}^{4}{\left[x+e^{2x} \right]dx}}$ is the area bounded by the four items:
   ♦ The curve y = f(x) = x+e^2x.
   ♦ The vertical line x = 0 (y-axis)
   ♦ The vertical line x = 4
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.9 below. Our task is to find the area of the violet shaded area.

Fig.23.9

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = 0 and b = 4

So $\small{h~=~\frac{b-a}{n}~=~\frac{4-0}{n}~=~\frac{4}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \left[f(a+\frac{4}{n})~+~f(a+\frac{8}{n})~+~f(a+\frac{12}{n})~+~~.~.~.~+f(a+(n)\frac{4}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \bigg[\big[a+\frac{4}{n}~+~e^{2(a+\frac{4}{n})}\big]~+~\big[a+\frac{8}{n}~+~e^{2(a+\frac{8}{n})}\big]~+~\big[a+\frac{12}{n}~+~e^{2(a+\frac{12}{n})}\big]~+~~.~.~.~}$

$\small{~~~~~~~~~.~.~.~+~\big[a+(n)\frac{4}{n}~+~e^{2(a+(n)\frac{4}{n})}\big]\bigg]}$

4. Let us determine the quantity inside the large square brackets. We have to do three summations:

(i) $\small{a~+~a~+~a~+~.~.~.~ \text{n terms}}$

$~=~\small{n a~=~n(0)~=~0}$

(ii) $\small{\frac{4}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{4}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{4(n+1)}{2}~=~2n~+~2}$

(Here we use the technique of arithmetic progression)

(iii) $\small{e^{2(a+\frac{4}{n})}~+~e^{2(a+\frac{8}{n})}~+~e^{2(a+\frac{12}{n})}~+~.~.~.~ \text{n terms}}$

$\small{~=~e^{2a}~\times~e^{\frac{8}{n}}~+~e^{2a}~\times~e^{\frac{16}{n}}~+~e^{2a}~\times~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms}}$

$\small{~=~e^{2a}\left(e^{\frac{8}{n}}~+~e^{\frac{16}{n}}~+~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms} \right)}$

$\small{~=~e^{\frac{8}{n}}~+~e^{\frac{16}{n}}~+~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms} }$

$\small{~~~~~~\because e^{2a}~=~e^{2(0)}~=~e^0~=~1}$

• So we have a geometric progression.

• First term = $\small{e^{\frac{8}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{16}{n}}}{e^{\frac{8}{n}}}~=~\frac{e^{\frac{24}{n}}}{e^{\frac{16}{n}}}~=~e^{\frac{8}{n}}}$

• Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^{n-1}~-~1 \right)}{r~-~1}}$

$\small{r^{n-1}~=~\left(e^{\frac{8}{n}}\right)^{n-1}~=~e^{\frac{8n-8}{n}}~=~e^{8-\frac{8}{n}}}$

• So we get:

Sum of all terms of the G.P

$\small{~=~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1}}$

$\small{r^{n-1}~=~\left(e^{\frac{8}{n}}\right)^{n-1}~=~e^{\frac{8n-8}{n}}~=~e^{8-\frac{8}{n}}}$

• So we get:

Sum of all terms inside the brackets

$\small{~=~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1}}$

5. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{4}{n} \big[0~+~2n\,+\,2~+~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \big[2n\,+\,2 \big]~+~\lim_{n\rightarrow \infty} \frac{4}{n} \big[\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~8~+~\lim_{n\rightarrow \infty} \frac{4}{n} \big[\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$

6. In the above expression, we need to evaluate the limit for the second term. It can be done as follows:

$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}}~\times~\left(\frac{e^8}{e^\frac{8}{n}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8~-~e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8 }{e^{\frac{8}{n}}~-~1} \big]~-~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8 }{e^{\frac{8}{n}}~-~1} \big]~-~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\frac{e^8}{2}\,\lim_{n\rightarrow \infty} \frac{8}{n} \big[\frac{1 }{e^{\frac{8}{n}}~-~1} \big]~-~\frac{1}{2}\,\lim_{n\rightarrow \infty} \frac{8}{n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\frac{e^8}{2}\,\lim_{n\rightarrow \infty}  \bigg[\frac{1 }{\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}} \bigg]~-~\frac{1}{2}\,\lim_{n\rightarrow \infty}  \big[\frac{e^{\frac{8}{n}} }{\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}} \big]}$

$\small{~=~\frac{e^8}{2}\,  \bigg[\frac{1}{1} \bigg]~-~\frac{1}{2}\, \big[\frac{1}{1} \big]~=~\frac{e^8~-~1}{2}}$

7. So the limit in (3) becomes:

$\small{8~+~\frac{e^8~-~1}{2}~=~\frac{16~+~e^8~-~1}{2}~=~\frac{15~+~e^8}{2}}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{8}{n}} \right]~=~e^{\frac{8}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{8}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$

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The link below gives a few more solved examples:

Exercise 23.8

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We have completed a discussion on the definite integral as limit of sum. In the next section, we will see fundamental theorem of calculus.

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