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

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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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23.21 - Definite Integrals

In the previous section, we completed a discussion on three standard integrals. In this section, we will see definite integrals.

Let us first see a method to calculate the area of a region. The region that we are considering here, is a special kind of region because, one of it's sides is not a straight line. The method can be written in 15 steps:

1. In fig.23.2 below, the red curve is the graph of y = f(x).

Fig.23.2

• We assume that, for all input x values, f(x) is non negative. So the graph is above the x-axis.
• There are two vertical lines: x = a and x = b. They are drawn in green color.
• So we have a region bounded by the four items below:
   ♦ The curve y = f(x)
   ♦ The vertical line x = a
   ♦ The vertical line x = b
   ♦ The horizontal line y = 0 (x-axis)
• Our aim is to find the area of this region.

2. In fig.23.3 below, the region is divided into strips.

Fig.23.3

• The strips have the following 4 properties:
(i) All strips are formed by vertical lines.
(ii) All strips have the same width. We will denote this width as 'h'.
(iii) There is a total of n strips.
(iv) Each of the n strips, has it's own position.
   ♦ The left edge of the first strip is at x = x₀ = a
   ♦ The left edge of the second strip is at x = x₁
   ♦ The left edge of the third strip is at x = x₂
   ♦ The left edge of the fourth strip is at x = x₃
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ The left edge of the r^th strip is at x = x_r−1
   ♦ The left edge of the (r+1)^th strip is at x = x_r
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ The left edge of the n^th strip is at x = x_n−1
   ♦ The right edge of the n^th strip is at x = x_n = b

3. Based on the above properties of the strips, we can write:
   ♦ x₀ = a
   ♦ x₁ = a+h
   ♦ x₂ = a+h+h = a+2h
   ♦ x₃ = a+2h+h = a+3h
   ♦ x₄ = a+3h+h = a+4h
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ x_r = a+rh
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ x_n = a+nh
4. Now we can write an expression for n.
• We have $\small{x_n~=~a+nh~=~b}$
$\small{\Rightarrow nh~=~b-a}$
$\small{\Rightarrow n~=~\frac{b-a}{h}}$

5. From the above expression, we get an important information:
• h is in the denominator. So when n increases, h decreases.
• Consequently, when n is close to infinity, h will be close to zero.

6. Now select any convenient strip as the r^th strip. In fig.23.3, it is named as ABCD.
• Note that, none of the strips is a rectangle. This is because, the top edge of any selected strip is not horizontal. Our present strip ABCD is also not a rectangle because, the top edge CD is not horizontal.

7. Since CD is not horizontal, we draw two new horizontal lines:
   ♦ Horizontal line CL through C
   ♦ Horizontal line DM through D
• Now we have three regions:
   ♦ Rectangle ABDM
   ♦ Region ABDCA
   ♦ Rectangle ABLC

8. Let us compare the areas of the three regions. From the fig., it is clear that:
• Area of region ABDCA is less than the area of the rectangle ABDM
• Area of region ABDCA is greater than the area of the rectangle ABLC

9. Consider the situation when n is close to infinity.
• In such a situation,
   ♦ h will be very close to zero.
   ♦ B will be very close to A
   ♦ L will be very close to C
   ♦ Segment MD will be very close to segment CL
• In such a situation, all three areas mentioned in (8), can be considered to be equal.

10. It follows that:
• When n is close to infinity, the area of region ABCDA can be calculated easily because, it is equal to the area of rectangle ABLC.
• Area of rectangle ABLC = AC  × AB = f(x_r-1)  × h
• Here we are using the lower rectangle ABLC to get the area of the region ABCDA. Length of the rectangle is left edge AC.

11. Similarly, it follows that:
• When n is close to infinity, the area of region ABCDA can be calculated easily because, it is equal to the area of rectangle ABDM.
• Area of rectangle ABDM = BD  × AB = f(x_r)  × h
• Here we are using the upper rectangle ABDM to get the area of the region ABCDA. Length of the rectangle is right edge BD.

12. The information in both (10) and (11) are applicable to all strips.
• Let us first apply the information in (10).
   ♦ Area of first strip = left edge × base = f(x₀) × h
   ♦ Area of second strip = left edge × base = f(x₁) × h
   ♦ Area of third strip = left edge × base = f(x₂) × h
   ♦ Area of fourth strip = left edge × base = f(x₃) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of r^th strip = left edge × base = f(x_r−1) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of n^th strip = left edge × base = f(x_n−1) × h
• So total area of all n strips
= h[f(x₀) + f(x₁) + f(x₂) + . . . +f(x_n−1)]= $\small{h\sum\limits_{r=0}^{r=n-1}{f(x_r)}}$
• Based on the result in (3), we can write x values in terms of a and h. We get:
Total area of all n strips
= h[f(x₀) + f(x₁) + f(x₂) + . . . +f(x_n−1)]
= h[f(a) + f(a+h) + f(a+2h) + . . . +f(a+(n−1)h)]
= $\small{h\sum\limits_{k=0}^{k=n-1}{f(a+kh)}}$
• When n is close to infinity, h will be close to zero. In such a situation, this sum will be equal to the area of PRSQP.
• That is.,
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a)~+~f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+(n-1)h) \right]}$

13. Let us now apply the information in (11).
   ♦ Area of first strip = right edge × base = f(x₁) × h
   ♦ Area of second strip = right edge × base = f(x₂) × h
   ♦ Area of third strip = right edge × base = f(x₃) × h
   ♦ Area of fourth strip = right edge × base = f(x₄) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of r^th strip = right edge × base = f(x_r) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of n^th strip = right edge × base = f(x_n) × h
• So total area of all n strips
= h[f(x₁) + f(x₂) + f(x₃) + . . . +f(x_n)] = $\small{h\sum\limits_{r=1}^{r=n}{f(x_r)}}$
• Based on the result in (3), we can write x values in terms of a and h. We get:
• Total area of all n strips
= h[f(x₁) + f(x₂) + f(x₃) + . . . +f(x_n)]
= h[f(a+h) + f(a+2h) + f(a+3h) + . . . +f(a+nh)]
= $\small{h\sum\limits_{k=1}^{k=n}{f(a+kh)}}$
• When n is close to infinity, h will be close to zero. In such a situation, this sum will be equal to the area of PRSQP.
• That is.,
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

14. Let us compare the methods.
• Step (12) gives us a method to find the area of PRSQP.
It uses the left edge of each rectangle.
• Step (13) also gives us a method to find the area of PRSQP.
It uses the right edge of each rectangle.
• We can use any one of the two methods. It is convenient to use the method in (13). This is because, it has f(a+nh) instead of f(a+(n−1)h)

15. So we can write:
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

• Using the result in step (4), we can replace the 'h' outside the square brackets. We get:
Area of PRSQP =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

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Let us see a solved example:
Solved Example 23.69
Find the area bounded by the four items:
   ♦ The curve y = f(x) = x².
   ♦ The vertical line x = 2
   ♦ The vertical line x = 3
   ♦ The horizontal line y = 0 (x-axis)
Solution:
1. A rough sketch is shown in the fig.23.4 below:

Fig.23.4

• Our task is to find the area of the violet shaded area.
2. We have the formula for area:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right]}$
• In our present case, a = 2 and b = 3
So $\small{h~=~\frac{b-a}{n}~=~\frac{3-2}{n}~=~\frac{1}{n}}$

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

= $\small{\lim_{n\rightarrow \infty} \frac{3-2}{n} \left[f(a+\frac{1}{n})~+~f(a+\frac{2}{n})~+~~.~.~.~+f(a+(n)\frac{1}{n}) \right]}$

= $\small{\lim_{n\rightarrow \infty} \frac{1}{n} \left[(a+\frac{1}{n})^2~+~(a+\frac{2}{n})^2~+~~.~.~.~+(a+\frac{n}{n})^2 \right]}$

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

$\small{ \left[\left(a^2 + \frac{2a}{n} + \frac{1^2}{n^2} \right)~+~\left(a^2 + \frac{4a}{n} + \frac{2^2}{n^2} \right)~+~.~.~.~+~\left(a^2 + \frac{2an}{n} + \frac{n^2}{n^2} \right) \right]}$

5. So we have to do three summations:

(i) $\small{a^2~+~a^2~+~a^2~+~.~.~.~ \text{n terms}~=~\small{n a^2~=~n(2^2)~=~4n}}$

$\small{~~~~~=~\small{n a^2~=~n(2^2)~=~4n}}$

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

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

(Here we use the technique of arithmetic progression)

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

$\small{~~~~~=~\frac{1}{n^2}\left(\frac{2n^3 + 3 n^2 + n}{6} \right)~=~\frac{n}{3}~+~\frac{1}{2}~+~\frac{1}{6n}}$

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

6. So the total of three summations is:

$\small{4n~+~n+1~+~\frac{n}{3}~+~\frac{1}{2}~+~\frac{1}{6n}}$

$\small{~=~\frac{2}{n}~+~\frac{19\,n}{3}-\frac{9}{2}+\frac{1}{6n}~=~\frac{13}{6n}~+~\frac{19\,n}{3}-\frac{9}{2}}$

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

$\small{\lim_{n\rightarrow \infty} \frac{1}{n} \left[\frac{13}{6n}~+~\frac{19\,n}{3}-\frac{9}{2} \right]}$

$\small{~=~\lim_{n\rightarrow \infty}  \left[\frac{13}{6n^2}~+~\frac{19}{3}-\frac{9}{2n} \right]~=~\frac{19}{3}}$

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Let us write the definition of definite integral. It can be written in 3 steps:
1. We have seen a method to find the area.
• Symbolically, this area is denoted as $\small{\int_a^b{\left[f(x) \right]dx}}$.
   ♦ This is the definite integral of f(x).
2. So we can write:
Area = $\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right] ~=~\int_a^b{\left[f(x) \right]dx}}$
3. Within the square brackets, we have a sum. We are taking the limiting value of that sum. So it is a limit of sum.
• Then we can define the definite integral $\small{\int_a^b{\left[f(x) \right]dx}}$, as the limit of sum.

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Now we will write an important point. It can be written in 3 steps:
1. In the above solved example, we calculated the definite integral of f(x) = x²
2. It is clear that the variable associated with the function f is x.
• But we applied the limit to the variable n.
3. We see that, the result will depend upon f(x), a and b. The result will not depend upon n.
• So n is called the variable of integration. It is also called the dummy variable.

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

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