Compute $\displaystyle \int^1_0 \int^{x^2}_0 \frac{y}{e^x} \, dy \, dx$

It's been a couple of years since I've done any real integration and we just started doing double integrals in my Calculus 3 class. I can't remember what do do from here:

$$\int^1_0 \left.\frac{1}{e^x}\frac{y^2}{2}\right|^{x^2}_0 \, dx = \frac{1}{2} \int^1_0 \frac{x^4}{e^x} \, dx$$

From here I assumed integration by parts:


$$du=e^x \, dx$$

$$dv=4x^3 \, dx$$


Setting this up I get:

$$\frac{e^x}{2} x^4-\int^1_0 x^4e^x \, dx$$

This is where I'm stuck. I'm not sure where to go from here.

  • 2
    $\begingroup$ Careful when integrating by parts. Your $dv$ should be $x^{4}dx$, with $v$ then being $\frac{x^{5}}{5}$. $\endgroup$ – Sinister Cutlass Nov 18 '15 at 0:43
  • $\begingroup$ Oh so I put the current function as the $dv$ and then its integral as the $v$? $\endgroup$ – hax0r_n_code Nov 18 '15 at 0:45
  • 1
    $\begingroup$ You usually want repeated derivatives of $u$ to go to 0 (polynomials). $\endgroup$ – Ben Longo Nov 18 '15 at 0:48

Let me suggest a different strategy. Take the integral $$\int x^ne^{-x}\mathrm{d}x$$ Take $$u=x^n,\quad\mathrm{d}u=nx^{n-1}\mathrm{d}x$$ $$\mathrm{d}v=e^{-x}\mathrm{d}x,\quad v=-e^{-x}$$ We then have $$\int x^ne^{-x}\mathrm{d}x=vu-\int v\mathrm{d}u=-x^ne^{-x}+\int nx^{n-1}e^{-x}\mathrm{d}x$$ Notice how this reduces the power $x$ is raised to.

Can you apply this here and whittle down the power until the only term you have to integrate is $$A\int e^{\pm x}\mathrm{d}x$$ for some constant $A$?

The reason I suggest this is that with your choice of $u$ and $\mathrm{d}v$, the power that $x$ is raised to will keep rising, which isn't helpful at all.


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