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.

---

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

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

---

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

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:

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

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

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

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

---

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

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) 

---

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

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.

---

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

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:

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

---

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

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

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

---

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

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

---

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.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

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:
$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right) \right]_{0}^{2}}$

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

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

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

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

---

The link below gives a few more solved examples:

Exercise 23.10

---

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

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

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:
$\small{I~=~F(\frac{\pi}{6})\,-\,F(\frac{\pi}{4})~=~\left[5x + \frac{5}{2} \log \left|x+1 \right| -\frac{45}{2} \log \left|x+3 \right| \right]_{1}^{2}}$

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

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

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

---

The link below gives a few more solved examples:

Exercise 23.9

---

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

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com