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CHAPTER 4 The Integral

b
that area is We conclude therefore that
a f (x) dx.
b
f (x) dx = F (b) ’ F (a).
a




y = f (x)




a b




F (b) _ F (a)



Fig. 4.10

Finally, if G is any other antiderivative for f then G(x) = F (x) + C. Hence
b
G(b) ’ G(a) = [F (b) + C] ’ [F (a) + C] = F (b) ’ F (a) = f (x) dx.
a
That is the content of the Fundamental Theorem of Calculus.

You Try It: Calculate the area below the curve y = ’x 2 + 2x + 4 and above the
x-axis.



4.3 Signed Area
Without saying so explicitly, we have implicitly assumed in our discussion of area in
the last section that our function f is positive, that is its graph lies about the x-axis.
But of course many functions do not share that property. We nevertheless would
like to be able to calculate areas determined by such functions, and to calculate the
corresponding integrals.
This turns out to be simple to do. Consider the function y = f (x) shown in
Fig. 4.11. It is negative on the interval [a, b] and positive on the interval [b, c].
112 CHAPTER 4 The Integral

Suppose that we wish to calculate the shaded area as in Fig. 4.12. We can do so by
breaking the problem into pieces.




Fig. 4.11




Fig. 4.12

Of course, because f ≥ 0, the area between x = b and x = c is given by the
c
integral b f (x) dx, just as we have discussed in the last section. But our discussions
do not apply directly to the area between x = a and x = b. What we can do is
instead consider the function g = ’f . Its graph is shown in Fig. 4.13. Of course
g is a positive function on [a, b], except at the endpoints a and b; and the area
under g”between x = a and x = b”is just the same as the shaded area between
113
CHAPTER 4 The Integral




Fig. 4.13


x = a and x = b in Fig. 4.14. That area is
b b
g(x) dx = ’ f (x) dx.
a a




Fig. 4.14

In total, the aggregate shaded area exhibited in Fig. 4.15, over the entire interval
[a, c], is
b c
’ f (x) dx + f (x) dx.
a b
114 CHAPTER 4 The Integral




Fig. 4.15

What we have learned is this: If f (x) < 0 on the interval under discussion, then
the integral of f will be a negative number. If we want to calculate positive area
then we must interject a minus sign.
Let us nail down our understanding of these ideas by considering some examples.
EXAMPLE 4.8
Calculate the (positive) area, between the graph of f (x) = x 3 ’ 2x 2 ’
11x + 12 and the x-axis, between x = ’3 and x = 4.
SOLUTION
Consider Fig. 4.16. It was drawn using the technique of Section 3.1, and it
plainly shows that f is positive on [’3, 1] and negative on [1, 4]. From the




Fig. 4.16
115
CHAPTER 4 The Integral

discussion preceding this example, we know then that
1 4
Area = f (x) dx ’ f (x) dx
’3 1
1
= x 3 ’ 2x 2 ’ 11x + 12 dx
’3
4
’ x 3 ’ 2x 2 ’ 11x + 12 dx
1
1
4 2x 3 11x 2
x
= ’ ’ + 12x
4 3 2 ’3
4
x 4 2x 3 11x 2
’ ’ ’ + 12x (—)
.
4 3 2 1
Here we are using the standard shorthand
F (x)|b
a

to stand for
F (b) ’ F (a).
Thus we have
160 297
(—) = + .
3 12
Notice that, by design, each component of the area has made a positive
contribution to the ¬nal answer. The total area is then
937
Area = .
12
EXAMPLE 4.9
Calculate the (positive) area between f (x) = sin x and the x-axis for
’2π ¤ x ¤ 2π.
SOLUTION
We observe that f (x) = sin x ≥ 0 for ’2π ¤ x ¤ ’π and 0 ¤ x ¤ π.
Likewise, f (x) = sin x ¤ 0 for ’π ¤ x ¤ 0 and π ¤ x ¤ 2π. As a result
’π 0
Area = sin x dx ’ sin x dx
’2π ’π

π
+ sin x dx ’ sin x dx.
0 π
116 CHAPTER 4 The Integral

This is easily calculated to equal
2 + 2 + 2 + 2 = 8.

You Try It: Calculate the (positive) area between y = x 3 ’ 6x 2 + 11x ’ 6 and
the x-axis.
EXAMPLE 4.10
Calculate the signed area between the graph of y = cos x + 1/2 and the
x-axis, ’π/2 ¤ x ¤ π .

SOLUTION
This is easy, because the solution we seek is just the value of the integral:
π 1
Area = cos x + dx
2
’π/2
π
x
= sin x +
2 ’π/2
’π
π
= 0+ ’ ’1 +
2 4

= + 1.
4
Math Note: In the last example, we have counted positive area as positive and
negative area as negative. Our calculation shows that the aggregate area is positive.
We encourage the reader to draw a graph to make this result plausible.

You Try It: Calculate the actual positive area between the graph of y = x 2 ’ 4,
’5 ¤ x ¤ 5 and the x-axis.

You Try It: Calculate the signed area between the graph of y = x 2 ’ 4 and the
x-axis, ’4 ¤ x ¤ 5.



4.4 The Area Between Two Curves
Frequently it is useful to ¬nd the area between two curves. See Fig. 4.17. Following
the model that we have set up earlier, we ¬rst note that the intersected region has
left endpoint at x = a and right endpoint at x = b. We partition the interval [a, b]
as shown in Fig. 4.18. Call the partition
P = {x0 , x1 , . . . , xk }.
117
CHAPTER 4 The Integral




Fig. 4.17




Fig. 4.18

Then, as usual, we erect rectangles over the intervals determined by the partition
(Fig. 4.19).
Notice that the upper curve, over the interval [a, b], is y = f (x) and the lower
curve is y = g(x) (Fig. 4.17). The sum of the areas of the rectangles is therefore
k
[f (xj ) ’ g(x)] · x.
j =1

But of course this is a Riemann sum for the integral
b
[f (x) ’ g(x)] dx.
a
118 CHAPTER 4 The Integral




Fig. 4.19




Y
We declare this integral to be the area determined by the two curves.
FL
EXAMPLE 4.11
Find the area between the curves y = x 2 ’ 2 and y = ’(x ’ 1)2 + 3.
AM

SOLUTION
We set the two equations equal and solve to ¬nd that the curves intersect
at x = ’1 and x = 2. The situation is shown in Fig. 4.20. Notice that y =
TE



’(x ’ 1)2 + 3 is the upper curve and y = x 2 ’ 2 is the lower curve. Thus the
desired area is
2
Area = [’(x ’ 1)2 + 3] ’ [x 2 ’ 2] dx
’1
2
= ’2x 2 + 2x + 4 dx
’1
2
’2x 3
= + x + 4x
2
3 ’1
’16 2
= +4+8 ’ +1’4
3 3
= 9.
The area of the region determined by the two parabolas is 9.
EXAMPLE 4.12
Find the area between y = sin x and y = cos x for π/4 ¤ x ¤ 5π/4.
119
CHAPTER 4 The Integral

_2
y = x2




_ (x _ 1)2 + 3
y=


Fig. 4.20

SOLUTION
On the given interval, sin x ≥ cos x. See Fig. 4.21. Thus the area we wish to
compute is
5π/4
Area = [sin x ’ cos x] dx
π/4
x=5π/4
= [’ cos x ’ sin x]x=π/4
√ √ √ √
2 2 2 2
=’’ ’’ ’’ ’
2 2 2 2

= 2 2.

y




y = sin x

x
y = cos x




Fig. 4.21


You Try It: Calculate the area between y = sin x and y = cos x, ’π ¤ x ¤ 2π.

You Try It: Calculate the area between y = x 2 and y = 3x + 4.
120 CHAPTER 4 The Integral

4.5 Rules of Integration
We have been using various rules of integration without enunciating them explicitly.
It is well to have them recorded for future reference.


4.5.1 LINEAR PROPERTIES
I. If f, g are continuous functions on [a, b] then
b b b
f (x) + g(x) dx = f (x) dx + g(x) dx.
a a a
II. If f is a continuous function on [a, b] and c is a constant then
b b
cf (x) dx = c f (x) dx.
a a



4.5.2 ADDITIVITY
III. If f is a continuous on [a, c] and a < b < c then
b c c
f (x) dx + f (x) dx = f (x) dx.
a b a

You Try It: Calculate
3
4x 3 ’ 3x 2 + 7x ’ 12 cos x dx.
1




Exercises
1. Calculate each of the following antiderivatives:
Antiderivative of x 2 ’ sin x
(a)
Antiderivative of e3x + x 4 ’ 2
(b)
Antiderivative of 3t 2 + (ln t/t)
(c)
Antiderivative of tan x ’ cos x + sin 3x
(d)
Antiderivative of cos 3x + sin 4x + 1
(e)
Antiderivative of (cos x) · esin x
(f)
121
CHAPTER 4 The Integral

2. Calculate each of the following inde¬nite integrals:
x sin x 2 dx
(a)
(3/x) ln x 2 dx
(b)
sin x · cos x dx
(c)
tan x · ln cos x dx
(d)
sec2 x · etan x dx
(e)
(2x + 1) · (x 2 + x + 7)43 dx
(f)
3. Use Riemann sums to calculate each of the following integrals:
22
1 x + x dx
(a)
1 2
(b) ’1 (’x /3) dx
4. Use the Fundamental Theorem of Calculus to evaluate each of the following
integrals:
32
1 x ’ 4x + 7 dx
3
(a)
2
6
xex ’ sin x cos x dx
(b) 2
4

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