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

432 Chapter 6 Applications of Integration
1026 2 1210
(b) The line through P 1 and P 2 has slope 18 2 10 − 2 184
8 − 223 and equation
N 2 1210 − 223st 2 10d &? N − 223t 1 1440. The area between f and this
line is
y 18
f f std 2 s223t 1 1440dg dt − y 18
s2t 3 1 20t 2 1 21t 1 23t 2 1440ddt
10
10
− y 18
s2t 3 1 20t 2 1 44t 2 1440ddt
10
−F2 t 4
4 1 20 t 3
3 1 44 t 2
18
G
2 2 1440t
− 26156 2 s28033 1 3 d < 1877
10
y
y=©
S¡
S S£
y=ƒ
0 a
b x
FIGURE 11
Thus the level of infectiousness for this patient is about 1877 (cellsymL) ? days.
If we are asked to find the area between the curves y − f sxd and y − tsxd where
f sxd > tsxd for some values of x but tsxd > f sxd for other values of x, then we split the
given region S into several regions S 1 , S 2 , . . . with areas A 1 , A 2 , . . . as shown in Figure
11. We then define the area of the region S to be the sum of the areas of the smaller
regions S 1 , S 2 , . . . , that is, A − A 1 1 A 2 1 ∙ ∙ ∙. Since
| f sxd 2 tsxd | H − f sxd 2 tsxd
tsxd 2 f sxd
when f sxd > tsxd
when tsxd > f sxd
n
we have the following expression for A.
3 The area between the curves y − f sxd and y − tsxd and between x − a and
x − b is
A − y b
a
| f sxd 2 tsxd | dx
When evaluating the integral in (3), however, we must still split it into integrals corresponding
to A 1 , A 2 , . . . .
Example 6 Find the area of the region bounded by the curves y − sin x, y − cos x,
x − 0, and x − y2.
SOLUtion The points of intersection occur when sin x − cos x, that is, when x − y4
(since 0 < x < y2). The region is sketched in Figure 12.
y
y =cos x
y=sin x
x=0
A¡
A
x= π 2
FIGURE 12
0 π π
x
4 2
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