James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 6.1 Areas Between Curves 429
y
In the case where both f and t are positive, you can see from Figure 3 why (2) is true:
S
y=ƒ
y=©
A − farea under y − f sxdg 2 farea under y − tsxdg
− y b
f sxd dx 2 y b
tsxd dx − y b
f f sxd 2 tsxdg dx
a
a
a
0 a
b x
FIGURE 3
A − y b
f sxddx 2 y b
tsxddx
a
a
y
y=´
x=1
y=x Îx
1
0
1
x
FIGURE 4
y
y T
y B
y T -y B
Îx
0 a
b
x
FIGURE 5
y T =2x-≈
y
(1, 1)
Example 1 Find the area of the region bounded above by y − e x , bounded below by
y − x, and bounded on the sides by x − 0 and x − 1.
SOLUtion The region is shown in Figure 4. The upper boundary curve is y − e x and
the lower boundary curve is y − x. So we use the area formula (2) with f sxd − e x ,
tsxd − x, a − 0, and b − 1:
A − y 1
se x 2 xd dx − e x 2 1 2 x 2 g 1 0
0
− e 2 1 2 2 1 − e 2 1.5 n
In Figure 4 we drew a typical approximating rectangle with width Dx as a reminder of
the procedure by which the area is defined in (1). In general, when we set up an integral
for an area, it’s helpful to sketch the region to identify the top curve y T , the bottom curve
y B , and a typical approximating rectangle as in Figure 5. Then the area of a typical rectangle
is sy T 2 y B d Dx and the equation
A − lim
n l ` o n
sy T 2 y B d Dx − y b
sy T 2 y B d dx
i−1
a
summarizes the procedure of adding (in a limiting sense) the areas of all the typical
rectangles.
Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Figure 3
the right-hand boundary reduces to a point. In the next example both of the side boundaries
reduce to a point, so the first step is to find a and b.
Example 2 Find the area of the region enclosed by the parabolas y − x 2 and
y − 2x 2 x 2 .
SOLUtion We first find the points of intersection of the parabolas by solving
their equations simultaneously. This gives x 2 − 2x 2 x 2 , or 2x 2 2 2x − 0. Thus
2xsx 2 1d − 0, so x − 0 or 1. The points of intersection are s0, 0d and s1, 1d.
We see from Figure 6 that the top and bottom boundaries are
y T − 2x 2 x 2 and y B − x 2
The area of a typical rectangle is
Îx
(0, 0)
y B =≈
x
sy T 2 y B d Dx − s2x 2 x 2 2 x 2 d Dx
and the region lies between x − 0 and x − 1. So the total area is
FIGURE 6
A − y 1
s2x 2 2x 2 d dx − 2 y 1
sx 2 x 2 d dx
0
0
− 2F x 2
2 2 x 3
1
3G0
− 2S 1 2 2 1 3D − 1 3
n
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