I am not sure if i am not understanding the concept of double integration right, or if i am just making a silly mistake, but here is the problem:
given the function $f(x,y)=24xy, y>0, x>0$ and $x+y<=1 $ fint the $E[X,Y]$
$E[X,Y] = \int \int 24x^2y^2dxdy$ =
$24 \int_0^xx^2 (\int_0^{1-x}y^2dy) dx $ =
$24 \int_0^xx^2 [\frac{y^3}{3}]_0^{1-x}dx $=
$24 \int_0^xx^2 \frac{(1-x)^3}{3}dx $=
$8 \int_0^xx^2 (1-x)^3dx $=
$8[\frac{x^3}{x} \frac{(1-x)^4}{4}\frac{-x^2}{2}]_0^1$
= $ 8-\frac {5}{12}$
However, my professor gave us different solution:
$E[X,Y] = \int \int 24x^2y^2dxdy$ =
$24 \int_0^xx^2 (\int_0^{1-x}y^2dy) dx $ =
$3 \int_0^1 x^2 (1-x)^3 dx = $
$ \frac{2}{15}$
What is the problem with my calculation?
Where does the 3 come from in $3 \int_0^1 x^2 (1-x)^3 dx $ ?
I believe you either misread '8' as '3', or there is a typo on your prof's part.
\begin{align} E &= \int_{x=0}^{x=1}\int_{y=0}^{y=1-x} 24x^2y^2 \, dy \, dx \\ &= 24\int_{x=0}^{x=1}x^2\int_{y=0}^{y=1-x} y^2 \, dy \,dx \\ &= 24\int_{x=0}^{x=1}x^2\left[\frac{y^3}{3}\right]_{y=0}^{y=1-x} \,dx \\ &= 8\int_{0}^{1}x^2(1-x)^3 \,dx \\ &= 8\int_{0}^{1}x^2(1-x)^3 \,dx \\ &= 8\int_{0}^{1}x^2(1-3x+3x^2-x^3) \,dx \\ &= 8\int_{0}^{1}x^2-3x^3+3x^4-x^5 \,dx \\ &= 8\left[\frac{x^3}{3}-\frac{3x^4}{4}+\frac{3x^5}{5}-\frac{x^6}{6}\right]_0^1\\ &= 8\left(\frac 13-\frac34+\frac35-\frac16\right)\\ &= 8\cdot\frac{1}{60} \\ &=\frac2{15} \end{align}