We are given the following function:

$$f(x, y)= \begin{Bmatrix} 5e^{x^2}\:\:\:\:y\leq x \\ 5e^{y^2}\:\:\:\: y> x \end{Bmatrix}$$

which is bounded by the rectangle $D =[0,\:9]$ x $[0,\:9]$ in the plane.

How do I evaluate $\int\int_D f(x, y)dA$?

I've tried to graph the functions, but I don't get anywhere without integrating $e^{x^2}$ which doesn't give me an exact solution. Both functions seems to be defined for all the whole $xy-$plane. (If the the line $y=x$ would divide $y>x$ and $y<x$)

This integral has had me struggling for several hours now, I would really appreciate some help!


Here are some suggestions. The domain of integration can be split along the diagonal y=x and you can calculate the double integrals separately. Since you would ideally like to get a term such as $x \exp{x^2}$ in order to solve the integral (you may note that integrating this wrt. $x$ would yield $\frac{5}{2}\exp{x^2}$) I would suggest first integrating the $5\exp{x^2}$ in the y direction and then evaluate at the limits $y=0$ and $y=x$ and then doing the same for the other term. Hope this helps!

  • $\begingroup$ Thanks for the answer! I tried to figure out when the graph would cross the y=9 line, and that would be at x=sqrt(ln(5/4). The problem however is that, even with evaluation the limits of y=0 and y=x, I still end up with integrating e^x^2, which was the problem in the first place. $\endgroup$ – Zack King Feb 24 at 20:14
  • $\begingroup$ Integrate $5\exp{x^2}$ between the limits $y=0$ and $y=x$ and $x=0$ and $x=9$. Integrate wrt. $y$ first. I believe that should give you an expression that is possible to integrate quite easily $\endgroup$ – Thomas Fjærvik Feb 24 at 20:16
  • $\begingroup$ I will try that, thank you very much! $\endgroup$ – Zack King Feb 24 at 20:20
  • $\begingroup$ No problem ! If it works, you can do almost exactly the same for $5\exp{y^2}$. Please accept answer if it turns out to help/be correct :) $\endgroup$ – Thomas Fjærvik Feb 24 at 20:22
  • 1
    $\begingroup$ Yes! Earlier when I graphed it, clearly the two areas were the same, so I followed your steps and by symmetry I could multiply it by two :)) $\endgroup$ – Zack King Feb 24 at 20:24

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