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multivariable_09_Applications_of_Integration_4up

# multivariable_09_Applications_of_Integration_4up - 170...

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9 Applications of Integration We have seen how integration can be used to find an area between a curve and the x -axis. With very little change we can find some areas between curves; indeed, the area between a curve and the x -axis may be interpreted as the area between the curve and a second “curve” with equation y = 0. In the simplest of cases, the idea is quite easy to understand. EXAMPLE 9.1 Find the area below f ( x ) = - x 2 + 4 x + 3 and above g ( x ) = - x 3 + 7 x 2 - 10 x + 5 over the interval 1 x 2. In figure 9.1 we show the two curves together, with the desired area shaded, then f alone with the area under f shaded, and then g alone with the area under g shaded. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 5 10 0 1 2 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 5 10 0 1 2 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 5 10 0 1 2 3 . Figure 9.1 Area between curves as a difference of areas. 169 170 Chapter 9 Applications of Integration It is clear from the figure that the area we want is the area under f minus the area under g , which is to say integraldisplay 2 1 f ( x ) dx - integraldisplay 2 1 g ( x ) dx = integraldisplay 2 1 f ( x ) - g ( x ) dx. It doesn’t matter whether we compute the two integrals on the left and then subtract or compute the single integral on the right. In this case, the latter is perhaps a bit easier: integraldisplay 2 1 f ( x ) - g ( x ) dx = integraldisplay 2 1 - x 2 + 4 x + 3 - ( - x 3 + 7 x 2 - 10 x + 5) dx = integraldisplay 2 1 x 3 - 8 x 2 + 14 x - 2 dx = x 4 4 - 8 x 3 3 + 7 x 2 - 2 x vextendsingle vextendsingle vextendsingle vextendsingle 2 1 = 16 4 - 64 3 + 28 - 4 - ( 1 4 - 8 3 + 7 - 2) = 23 - 56 3 - 1 4 = 49 12 . It is worth examining this problem a bit more. We have seen one way to look at it, by viewing the desired area as a big area minus a small area, which leads naturally to the difference between two integrals. But it is instructive to consider how we might find the desired area directly. We can approximate the area by dividing the area into thin sections and approximating the area of each section by a rectangle, as indicated in figure 9.2. The area of a typical rectangle is Δ x ( f ( x i ) - g ( x i )), so the total area is approximately n 1 summationdisplay i =0 ( f ( x i ) - g ( x i ))Δ x. This is exactly the sort of sum that turns into an integral in the limit, namely the integral integraldisplay 2 1 f ( x ) - g ( x ) dx. Of course, this is the integral we actually computed above, but we have now arrived at it directly rather than as a modification of the difference between two other integrals. In that example it really doesn’t matter which approach we take, but in some cases this second approach is better. 9.1 Area between curves 171 0 5 10 0 1 2 3 . . Figure 9.2 Area between curves. EXAMPLE 9.2 Find the area below f ( x ) = - x 2 + 4 x + 1 and above g ( x ) = - x 3 + 7 x 2 - 10 x + 3 over the interval 1 x 2; these are the same curves as before but lowered by 2. In figure 9.3 we show the two curves together. Note that the lower curve now dips below the x -axis. This makes it somewhat tricky to view the desired area as a big area minus a smaller area, but it is just as easy as before to think of approximating the area by rectangles. The height of a typical rectangle will still be f ( x i ) - g ( x i ), even if g ( x i ) is negative. Thus the area is integraldisplay 2 1 - x 2 + 4 x + 1 - ( - x 3 + 7 x 2 -

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