1
$\begingroup$

I am having trouble finding a way to solve this: $$\int^{1}_{0}\int^{1}_{\sqrt{y}}e^{x^3}dxdy$$

The solution is $\frac1 3[e-1]$.

I reduced the integral to $\int^{1}_{0}[e^{x^3}-x^2e^{x^3}]dx$ but I can't seem to solve it further. I tried substituition and integration by parts and I don't think Trig substitution will work, will it?

Also because this is a solution from a multivariate book(Marsden,Tromba Vector Calculus) it should be solved without the use of the $\Gamma$ function.

Thanks in advance.

  • 0
    $$\int_0^1x^2e^{x^3}\ dx=\frac13\int_0^1e^u\ du$$ – 2017-01-22
  • 0
    @SimplyBeautifulArt Yes but that leaves the integral of $e^{x^3}$ which is apparently non elementary so I would guess that I can't separate them using additivity?? – 2017-01-22

1 Answers 1

7

Just switch order by drawing the region out,

$$\int_{0}^{1} \int_{\sqrt{y}}^{1} e^{x^3}dx dy= \int_{0}^{1} \int_{0}^{x^2} e^{x^3}dy dx$$

  • 3
    When it's so trivial you die when you see the answer. – 2017-01-22
  • 1
    I see my mistake I had $x^2$ in the lower boundary of the inner interval. Thanks – 2017-01-22
  • 0
    when you see double integrals, try swappin' 'em. – 2017-01-23