I have the following question on my assignment, and I have no idea how to take this integral. I tried using the online integral calculator, but it says the function is unintegratable (yet it spits out this number: $3.679137994987764$, which my homework says is wrong, and I don't understand how it got). However, my homework insists there is an answer.
The problem is $$\int_{-1}^1 \int_{-4|x|}^{|x|} e^{x+y} \, dy \, dx.$$
I was able to do the internal integral, and get
$$\int_{-1}^1 e^{x+|x|}-e^{x-4|x|} \, dx$$
However, I have no idea what to do from here. Do I split the integral up into two pieces (one from $-1$ to $0$, and the other from $0$ to one), and treat this problem in a similar manner to $\int|x| \,dx$?
Edit: I tried solving it like Zain Patel's answer demonstrates, but the program isn't accepting his answer, nor the one I get.
Breaking the integral up, we get: $$\int_{-1}^{0}e^{x-x}-e^{5x}dx+\int_{0}^{1}e^{x+x}-e^{-3x}dx$$
Taking the integral, I get the following, which is different than Zain's answer. $$x-\frac{1}{5}e^{5x}|_{-1}^{0}+\frac{1}{2}e^{2x}-\frac{1}{3}e^{-3x}|_{0}^{1}$$
Solving, I get $$\frac{-1}{30}+\frac{1}{5e^5}+\frac{e^2}{2}-\frac{1}{3e^3}$$
Does anyone see where a potential error is?
Edit 2: Tony K pointed out that my last integral was incorrect. It should be positive, not negative.
$$\frac{-1}{30}+\frac{1}{5e^5}+\frac{e^2}{2}+\frac{1}{3e^3}$$
Precisely! We have $$\begin{align}\int_{-1}^1 e^{x+|x|} - e^{x - 4|x|} \, \mathrm{d}x &= \int_{-1}^0 e^{x \color{red}{-}x}-e^{x\color{red}{+}4x} \, \mathrm{d}x + \int_0^1 e^{x\color{green}{+}x} - e^{x\color{green}{-}4x} \, \mathrm{d}x \\ & = \int_{-1}^0 1 - e^{5x} \, \mathrm{d}x+\int_0^1 e^{2x} -e^{-3x} \, \mathrm{d}x \\ & = \bigg[x-\frac{1}{5}e^{5x}\bigg]_{-1}^0 + \bigg[\frac{1}{2}e^{2x} + \frac{1}{3}e^{-3x}\bigg]_0^1 \\ & = \frac{4}{5}-\frac{e^{-5}}{5} + \frac{e^2}{2} + \frac{e^{-3}}{3} - \frac{5}{6}\end{align}$$