Unformatted text preview: 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 xaxis. 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 Z 2 1x 2 + 4 x + 1(x 3 + 7 x 210 x + 3) dx = Z 2 1 x 38 x 2 + 14 x2 dx. This is of course the same integral as before, because the region between the curves is identical to the former region—it has just been moved down by 2. EXAMPLE 9.3 Find the area between f ( x ) =x 2 + 4 x and g ( x ) = x 26 x + 5 over the interval 0 ≤ x ≤ 1; the curves are shown in ﬁgure 9.4 . Generally we should interpret...
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 Fall '07
 JonathanRogawski
 Math, Calculus, ......

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