Question

If the joint probability density of two random variables is given by
f(x1,x2)=6e^(-2x1-3x2) for x1>0,X2>0
find the probabilities that the first random variable will take on a value between 1 and 2 and the second random variable will take on a value between 2 and 3.

Answer 1

http://www.wolframalpha.com/input/?i=integrate+6e%5E%28-2x-3y%29+dxdy+from+x%3D1+to+2+and+y%3D2+to+3

I've replaced x1 and x2 two with x and y. I this answer this the same one that my book gets.

Answer 2

But my book goes into this detail instead of the way wolfram doesL
\[\int\limits_{1}^{2}\int\limits_{2}^{3}6e^{-2x1-3x2}dx1dx2=(e^{-2}-e^{-4})(e^{-6}-e^{-9})\]

Answer 3

\[(e^{−2}−e^{−4})(e^{−6}−e^{−9})\]
How are they getting this part of the answer?

Answer 4

\(\large \int\limits_{1}^{2}\int\limits_{2}^{3}6e^{-2x1-3x2}dx1dx2\\ \large =6\int\limits_{1}^{2}e^{-2x_1}dx_1\int\limits_{2}^{3}e^{-3x_2}dx_2 \)

Answer 5

which actually gives you \((e^{−2}−e^{−4})(e^{−6}−e^{−9})\)
just separated the variables, because it could be separated, hence separated the integrals.