Page 131 - Calculus Demystified

118 CHAPTER 4 The Integral

Fig. 4.19

We declare this integral to be the area determined by the two curves.

EXAMPLE 4.11
Find the area between the curves y = x − 2 and y =−(x − 1) + 3.
2
2
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SOLUTION
We set the two equations equal and solve to find that the curves intersect
at x =−1 and x = 2. The situation is shown in Fig. 4.20. Notice that y =
2
2
−(x − 1) + 3 is the upper curve and y = x − 2 is the lower curve. Thus the
desired area is
2
Area = [−(x − 1) + 3]−[x − 2] dx
2
2
−1
2 2
= −1 −2x + 2x + 4 dx
−2x 3 2
2
= 3 + x + 4x
−1

= −16 + 4 + 8 − 2 + 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.

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