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 |
$\small{\int_{-2}^3 {\left[0.1x^3~-~1.6 \right]dx}}$
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.
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.
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]
Let us see a solved example:
Solved Example 23.75
Evaluate the definite integral $\small{\int_{-1}^1{\left[x+1 \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^1{\left[x+1 \right]dx}}$
•
First we find the indefinite integral F. We have:
$\small{F~=~\int{\left[x+1 \right]dx}~=~\frac{x^2}{2}\,+\,x}$
2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(-1)~=~\left[\frac{x^2}{2}\,+\,x \right]_{-1}^1}$
$\small{~=~\left[\frac{1^2}{2}\,+\,1 \right]~-~\left[\frac{(-1)^2}{2}\,+\,(-1) \right]}$
$\small{~=~\left[\frac{3}{2} \right]~-~\left[\frac{-1}{2} \right]~=~2}$
3. A check can be done in 3 steps:
(i) In fig.23.13 below, the triangle is filled with blue color.
 |
| Fig.23.13 |
• The area of the triangle will be equal to the required definite integral.
(ii) The triangle has a base of 2 units.
♦ The red line has the equation y = x+1
♦ The right side green line has the equation x = 1
• Solving the two equations, we get:x = 1, y = 2
• So height of the triangle is 2 units.
(iii) Then the area of the triangle = $\small{\frac{1}{2}(2)(2)~=~2~ \text{square units}}$
Solved Example 23.76
Evaluate the definite integral $\small{\int_{2}^3{\left[\frac{1}{x} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{2}^3{\left[\frac{1}{x} \right]dx}}$
•
First we find the indefinite integral F. We have:
$\small{F~=~\int{\left[\frac{1}{x} \right]dx}~=~\log x}$
2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(3)\,-\,F(2)~=~\left[\log x \right]_{2}^3}$
$\small{~=~\left[\log 3 \right]~-~\left[\log 2 \right]}$
$\small{~=~\log \left(\frac{3}{2} \right)}$
3. The required definite integral is equal to the area of the blue region in fig.23.14 below:
 |
| Fig.23.14 |
Solved Example 23.77
Evaluate the definite integral $\small{\int_{1}^2{\left[4x^3 - 5x^2 + 6x + 9 \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{1}^2{\left[4x^3 - 5x^2 + 6x + 9 \right]dx}}$
• First we find the indefinite integral F.
We have:
$\small{F~=~\int{\left[4x^3 - 5x^2 + 6x + 9 \right]dx}}$
$\small{~=~x^4 - \frac{5x^3}{3} + 3x^2 + 9x}$
2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(1)~=~\left[x^4 - \frac{5x^3}{3} + 3x^2 + 9x \right]_{1}^2}$
$\small{~=~\left[2^4 - \frac{(5)2^3}{3} + (3)2^2 + (9)2 \right]~-~\left[1^4 - \frac{(5)1^3}{3} + (3)1^2 + (9)1 \right]}$
$\small{~=~\left[16 - \frac{40}{3} + 12 + 18 \right]~-~\left[1 - \frac{5}{3} + 3 + 9 \right]}$
$\small{~=~\left[\frac{98}{3} \right]~-~\left[\frac{34}{3} \right]}$
$\small{~=~\frac{64}{3}}$
3. The required definite integral is equal to the area of the blue region in fig.23.15 below:
 |
| Fig.23.15 |
Solved Example 23.78
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{4}}{\left[\sin(2x) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\sin(2x) \right]dx}}$
• First we find the indefinite integral F.
We have:
$\small{F~=~\int{\left[\sin(2x) \right]dx}~=~\frac{-\cos(2x)}{2}}$
2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\frac{-\cos(2x)}{2} \right]_{0}^{\frac{\pi}{4}}}$
$\small{~=~\left[\frac{-\cos(\frac{\pi}{2})}{2} \right]~-~\left[\frac{-\cos(0)}{2} \right]}$
$\small{~=~\left[0 \right]~+~\left[\frac{1}{2} \right]}$
$\small{~=~\frac{1}{2}}$
3. The required definite integral is equal to the area of the blue region in fig.23.16 below:
 |
| Fig.23.16 |
Solved Example 23.79
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{2}}{\left[\cos(2x) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\cos(2x) \right]dx}}$
• First we find the indefinite integral F.
We have:
$\small{F~=~\int{\left[\cos(2x) \right]dx}~=~\frac{\sin(2x)}{2}}$
2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{2})\,-\,F(0)~=~\left[\frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}}}$
$\small{~=~\left[\frac{\sin(\pi)}{2} \right]~-~\left[\frac{\sin(0)}{2} \right]}$
$\small{~=~\left[0 \right]~-~\left[0 \right]}$
$\small{~=~0}$
3. The required definite integral is equal to the area of the blue region in fig.23.17 below:
 |
| Fig.23.17 |
The upper blue region and lower blue region, have the same area. But they are opposite in sign. So the net area is zero.
Solved Example 23.80
Evaluate the definite integral $\small{\int_{4}^{5}{\left[e^x \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{4}^{5}{\left[e^x \right]dx}}$
• First we find the indefinite integral F.
We have:
$\small{F~=~\int{\left[e^x \right]dx}~=~e^x}$
2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(5)\,-\,F(4)~=~\left[e^x \right]_{4}^{5}}$
$\small{~=~\left[e^5 \right]~-~\left[e^4 \right]}$
$\small{~=~e^4(e-1)}$
3. The required definite integral is equal to the area of the blue region in fig.23.18 below:
 |
| Fig.23.18 |
Solved Example 23.81
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{4}}{\left[\tan(x) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\tan(x) \right]dx}}$
• First we find the indefinite integral F.
We have:
$\small{F~=~\int{\left[\tan(x) \right]dx}~=~\log \left| \sec x\right|}$
2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\log \left| \sec x\right| \right]_{0}^{\frac{\pi}{4}}}$
$\small{~=~\left[\log \left| \sec (\frac{\pi}{4})\right| \right]~-~\left[\log \left| \sec (0)\right| \right]}$
$\small{~=~\left[\log \left|\sqrt{2} \right| \right]~-~\left[\log \left|1 \right| \right]~=~\log \left|\frac{\sqrt 2}{1} \right|~=~\log \left|\sqrt 2 \right|}$
$\small{~=~\frac{1}{2} \log \left|2 \right|}$
3. The required definite integral is equal to the area of the blue region in fig.23.19 below:
 |
| Fig.23.19 |
In the next section, we will see a few more solved examples.
Previous
Contents
Copyright©2025 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment