The integral on 0 to 8 of g(u)du = 15

What is the integral on 0 to 64 of g(t/8)?

What is the integral on 0 to 8 of g(8-t)?

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- May 4th 2009, 11:14 PMAngloSaxonProtestantU-sub definite integral
The integral on 0 to 8 of g(u)du = 15

What is the integral on 0 to 64 of g(t/8)?

What is the integral on 0 to 8 of g(8-t)? - May 4th 2009, 11:20 PMTheEmptySet
$\displaystyle \int_{0}^{64}g\left(\frac{t}{8} \right)dt$

let $\displaystyle u=\frac{t}{8} \implies du=\frac{1}{8}dt \iff 8du=dt$

Now we need to change the limits of integration$\displaystyle u =\frac{t}{8}$

When t=0 we get $\displaystyle u =\frac{0}{8}=0$

When t=64 we get $\displaystyle u =\frac{64}{8}=8$

Now we sub in to get

$\displaystyle \int_{0}^{64}g\left(\frac{t}{8} \right)dt=\int_{0}^{8}g(u)(8du)=8\int_{0}^{8}g(u)d u=8\cdot 15=120$

Try the same thing on the next one.

Good luck