121616949-math.192 -

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Unformatted text preview: 178 Chapter 9 Applications of Integration 10 5 .. ... ... . . .. .. ... ... . . . ... ... ... . . ....................................................... ........................................................ ... ............. ..... ..................................................................................... .. ................................................................................................................................................................................... ..... . . . . .... ........................................................... .. ..... ................................................................................................................................................................ . . ... . . . . . . . . . . . . . . . . . . ..... ........................................................................................... ... ....... . . . . . . . . . . . . . . . . ... ..... . 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.............................................................................. 0 0 1 2 3 Figure 9.1 0 1 2 5 .. ... ... . . .. .. ... ... . . . ... ... ... . . . ... ... ... . . . ... ... .... ... ............... . . ... .............. ... .................. ... ................................ ... ........................................................ . . . ... ........ .............................................. ... ................................................................. ... ............................................................... ... .................................................................... . ... . ..... . . . ............................................................................................................................................................. .............................................................................. ....................................... .................................................................................................. .............................................................................. 0 3 0 1 2 3 Area between curves as a difference of areas. 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: Z Z 2 2 f (x) − g(x) dx = 1 −x2 + 4x + 3 − (−x3 + 7x2 − 10x + 5) dx 1 Z = 2 x3 − 8x2 + 14x − 2 dx 1 ¯2 ¯ x4 8x3 2 = − + 7x − 2x¯¯ 4 3 1 1 8 16 64 − + 28 − 4 − ( − + 7 − 2) 4 3 4 3 56 1 49 = 23 − − = . 3 4 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 ∆x(f (xi ) − g(xi )), so the total area is approximately n−1 X i=0 (f (xi ) − g(xi ))∆x. ...
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