# How do I solve this double integral?

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)

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

## 1 Answer

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!

• 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. – Zack King Feb 24 at 20:14
• 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 – Thomas Fjærvik Feb 24 at 20:16
• I will try that, thank you very much! – Zack King Feb 24 at 20:20
• 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 :) – Thomas Fjærvik Feb 24 at 20:22
• 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 :)) – Zack King Feb 24 at 20:24