Breakdown of basic multivariable calculus help I am just starting on multi-variable calculus now and I am a bit stuck on a (very) basic explanation for a question:
$$  \int\limits_0 ^\infty \int\limits_y^\infty 6e^{-(2x+3y)} dx\ dy$$
$$  = \int\limits_0^\infty 3e^{-5y}\ dy$$
$$=3/5$$
I don't understand how we jumped from line 1 to line 2, and if I tried to integrate on X I could only reach this $-3 e^{-2x}\ dx$. 
So perhaps anyone can show the detailed breakdown of the equation above. 
 A: First, as I believe you're already aware, you evaluate each part from the inside out, i.e., you have
$$\int\limits_0 ^\infty \int\limits_y^\infty 6e^{-(2x+3y)} dx\ dy = \int\limits_0 ^\infty\left(\int\limits_y^\infty 6e^{-(2x+3y)} dx\right) dy \tag{1}\label{eq1A}$$
Inside those brackets, you treat $y$ as being a constant as you're only integrating wrt $x$. Thus, you get
$$\begin{equation}\begin{aligned}
\int\limits_y^\infty 6e^{-(2x+3y)} dx & = \left. 6\left(-\frac{1}{2}\right)e^{-(2x+3y)}\; \right\rvert_{y}^{\infty} \\
& = 0 - (-3)e^{-(2y+3y)} \\
& = 3e^{-5y}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note with the integration, you get $-3e^{-2x-3y}$, not $-3e^{-2x}$ as you stated, i.e., you should not have dropped the $-3y$ part. For the outer integration, you have
$$\begin{equation}\begin{aligned}
\int_{0}^{\infty}3e^{-5y} & = \left. 3\left(-\frac{1}{5}\right)e^{-5y}\; \right\rvert_{0}^{\infty} \\
& = 0 - \left(-\frac{3}{5}\right)e^{0} \\
& = \frac{3}{5}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
