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Unformatted text preview: Example
Evaluate the deﬁnite integral
6x 3 0 √
−1 5x4 + 12 dx. Solution: Let
u = 5x4 + 12 =⇒ du = 20x3 dx
so that 1
du.
2
The lower and upper limits of integration for x variable are
x3 dx = lower limit for x = −1, upper limit for x = 0. To convert this to a corresponding range of integration for the uvariable, we note that
u(−1) = 5(−1)4 + 12 = 17, u(0) = 5 · 04 + 12 = 12, so that the lower and upper limits for the u variable are
lower limit for u = 17, upper limit for u = 12. Although the limits of integration for x are in the usual order, i.e., −1 < 0, these need not be
true for the limits of integration for u; in fact, in this case the lower limit of integration for u
is larger than its upper limit. Note: One should not switch the limits of integration for the
u variable without also multiplying the resulting integral by (−1) !
Making the above substitutions,
0 6x3
√
dx = 6
3x4 + 12 0 1
√
· x3 dx
−1
−1
3x4 + 12
12 1
1
√·
=6
du
u 20
17
3 12 −1/2
=
u
du
10 −1
12
3 u1/2
=
10 1/2 17
√
3√
=
12 − 17
5
√
3√
2 3 − 17 ,
=
5
which is the ﬁnal answer. There is no need to return to the original variable of integration (in
this case, x) for a deﬁnite integral problem. ...
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus, Limits

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