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.

  • 3
    $\begingroup$ Are you sure you need to change the limits at all? Did you solve the inner integral? Recall that $\int_{0}^t e^{ax} \,\mathrm{d}x = \left. \frac{e^{ax}}{a}\right|_0^t = \frac{e^{at} -1 }{a}$. In your case, feeding in $a = y^2, t = y^5$ yields a form that is very easy to deal with by substitution. $\endgroup$ Oct 26, 2017 at 2:26
  • $\begingroup$ Are you just trying to evaluate the integral, or is this an exercise on bounds of integration? $\endgroup$ Oct 26, 2017 at 2:33
  • $\begingroup$ @stochasticboy321, I got so involved in changing the limits that I haven't considered anything else. Thanks for the insight. $\endgroup$
    – MilTom
    Oct 27, 2017 at 19:57

3 Answers 3


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.


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