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Chapter 6 | Applications of Integration
Let’s revisit Example 6.4 , only this time let’s integrate with respect to y . Let R be the region depicted in Figure 6.10 . Find the area of R by integrating with respect to y .
Figure 6.10 The area of region R can be calculated using one integral only when the curves are treated as functions of y .
Solution We must first express the graphs as functions of y . As we saw at the beginning of this section, the curve on the left can be represented by the function x = v ( y ) = y , and the curve on the right can be represented by the function x = u ( y ) =2− y . Now we have to determine the limits of integration. The region is bounded below by the x -axis, so the lower limit of integration is y =0. The upper limit of integration is determined by the point where the two graphs intersect, which is the point (1, 1), so the upper limit of integration is y =1. Thus, we have ⎡ ⎣ c , d ⎤ ⎦ = [0, 1]. Calculating the area of the region, we get A = ∫ c d ⎡ ⎣ u ( y )− v ( y ) ⎤ ⎦ dy = ∫ ⎡ ⎣ ⎛ ⎝ 2− y ⎞ ⎠ − y ⎤ ⎦ dy = y 2 y 3/2
⎤ ⎦ ⎥ |
⎡ ⎣ ⎢ 2 y −
0 1
0 1
2 3
2 −
= 5 6 . The area of the region is 5/6 units 2 .
6.5 Let’s revisit the checkpoint associated with Example 6.4 , only this time, let’s integrate with respect to y . Let R be the region depicted in the following figure. Find the area of R by integrating with respect to y .
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