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)?
$\displaystyle \int_{0}^{64}g\left(\frac{t}{8} \right)dt$
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