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. $\endgroup$ – stochasticboy321 Oct 26 '17 at 2:26
  • $\begingroup$ Are you just trying to evaluate the integral, or is this an exercise on bounds of integration? $\endgroup$ – Michael Hardy Oct 26 '17 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 '17 at 19:57

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.