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).
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| Fig.23.2 |
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We assume that, for all input x values, f(x) is non negative. So the graph is above the x-axis.
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There are two vertical lines: x = a and x = b. They are drawn in green color.
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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)
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Our aim is to find the area of this region.
2. In fig.23.3 below, the region is divided into strips.
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| Fig.23.3 |
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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 rth strip is at x = xr−1
♦ The left edge of the (r+1)th strip is at x = xr
♦ - - - -
♦ - - - -
♦ The left edge of the nth strip is at x = xn−1
♦ The right edge of the nth strip is at x = xn = 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
♦ - - - -
♦ - - - -
♦ xr = a+rh
♦ - - - -
♦ - - - -
♦ xn = a+nh
4. Now we can write an expression for n.
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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:
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h is in the denominator. So when n increases, h decreases.
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Consequently, when n is close to infinity, h will be close to zero.
6. Now select any convenient strip as the rth strip. In fig.23.3, it is named as ABCD.
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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
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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:
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Area of region ABDCA is less than the area of the rectangle ABDM
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Area of region ABDCA is greater than the area of the rectangle ABLC
9. Consider the situation when n is close to infinity.
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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
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In such a situation, all three areas mentioned in (8), can be considered to be equal.
10. It follows that:
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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.
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Area of rectangle ABLC = AC × AB = f(xr-1) × h
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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:
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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.
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Area of rectangle ABDM = BD × AB = f(xr) × h
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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.
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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 rth strip = left edge × base = f(xr−1) × h
♦ - - - -
♦ - - - -
♦ Area of nth strip = left edge × base = f(xn−1) × h
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So total area of all n strips
= h[f(x0) + f(x1) + f(x2) + . . . +f(xn−1)]= $\small{h\sum\limits_{r=0}^{r=n-1}{f(x_r)}}$
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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(xn−1)]
= h[f(a) + f(a+h) + f(a+2h) + . . . +f(a+(n−1)h)]
= $\small{h\sum\limits_{k=0}^{k=n-1}{f(a+kh)}}$
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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.
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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 rth strip = right edge × base = f(xr) × h
♦ - - - -
♦ - - - -
♦ Area of nth strip = right edge × base = f(xn) × h
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So total area of all n strips
= h[f(x1) + f(x2) + f(x3) + . . . +f(xn)] = $\small{h\sum\limits_{r=1}^{r=n}{f(x_r)}}$
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Based on the result in (3), we can write x values in terms of a and h. We get:
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Total area of all n strips
= h[f(x1) + f(x2) + f(x3) + . . . +f(xn)]
= h[f(a+h) + f(a+2h) + f(a+3h) + . . . +f(a+nh)]
= $\small{h\sum\limits_{k=1}^{k=n}{f(a+kh)}}$
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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.
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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.
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Step (12) gives us a method to find the area of PRSQP.
It uses the left edge of each rectangle.
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Step (13) also gives us a method to find the area of PRSQP.
It uses the right edge of each rectangle.
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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 = $\small{\lim_{h\rightarrow 0} h\left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$
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Using the result in step (4), we can replace the 'h' outside the square brackets. We get:
Area of PRSQP =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$
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:
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| Fig.23.4 |
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Our task is to find the area of the violet shaded area.
2. We have the formula for area:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right]}$
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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:
$\small{\lim_{n\rightarrow \infty} \frac{1}{n} \left[\frac{13}{6n}~+~\frac{19\,n}{3}-\frac{9}{2} \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \left[\frac{13}{6n^2}~+~\frac{19}{3}-\frac{9}{2n} \right]~=~\frac{19}{3}}$
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.
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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.
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Then we can define the definite integral $\small{\int_a^b{\left[f(x) \right]dx}}$, as the limit of sum.
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.
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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.
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So n is called the variable of integration. It is also called the dummy variable.
In the next section, we will see a few more solved examples.
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