I have the following integral:
$$\int_0^1 \int_0^{y^5} 7y^8 e^{xy^2} \, dx \, dy$$
I tried drawing a picture and finding new limits of integration but the integral was still difficult to solve which means my new limits were wrong.
Help is appreciated.
Integrating with respect to x gives us: $$ \int_0^1 7y^8\cdot y^{-2}e^{xy^2} dy |_{x=0} ^{x=y^5} $$
$$ \int_0^1 7y^6 (e^{y^5\cdot y^2} - 1) dy $$
$$ \int_0^1 7y^6 (e^{y^7} - 1) dy $$
And from there you find the answer by a simple subsitution of $u = y^7$
$$ \int_0^1 (e^{u} - 1) du $$
$$ (e^{y^7} - y^7) |_{y=0}^{y=1} $$ $$ = e-2 $$
$$ \int_0^{y^5} 7y^8 e^{xy^2} \, dx = \left. 7y^8 \frac {e^{xy^2}}{y^2} \right|_{x\,:=\,0}^{x\,:=\,y^5} = 7y^6 e^{y^7} - 7y^6. $$ $$ \int_0^1 e^{y^7} \big( 7y^6 \, dy \big) = \int_0^1 e^u\,du, \qquad \int_0^1 7y^6\,dy = 1. $$ The bounds in that first integral do not change because when $y=0,1$ respectively then $u=0,1.$
Doing this iterated integral means performing the integration in $x$ first and treating $y$ as a constant and then doing the integral for $y$. Notice, that you can pull the $7y^8$ out of the $x$ integral and do a $u-$substitution with what remains($u=xy^2$). You can change the boundary conditions from $x$ to $u$. After this step is complete you have another $u-$sub left to do.
