double integral: int [0,1] int [0,y^2] (3y^3 e^xy) dx dy

[mindy88]

it's a double integration problem, i don't know how to make all the symbols and such so this might be a little bit confusing to read

it's the integral from 0 to 1, the integral from 0 to y^2 of (3y^3 e^xy) dx dy

so the integral from 0 to y^2 of (3y^3 e^xy) dx
i set u=xy
(1/y)du=dx
after i plug in 0 and y^2, i get 3y^2 * e^(y^3)

then it's the integral from 0 to 1 of (3y^2 * e^(y^3)) dy
I set u=y^3
du=3y^2 dy
then i plug in 0 and 1 and i get e-1 as the answer

the book says the answer is e-2

i double checked my work but i can't find my mistake

 

[pka]

\(\displaystyle \L\begin{array}{l}
\int\limits_0^1 {\int\limits_0^{y^2 } {3y^3 e^{xy} dxdy} } \\
\int\limits_0^{y^2 } {3y^3 e^{xy} dx = \left. {3y^2 e^{xy} } \right|_0^{y^2 } = 3y^2 e^{y^3 } - 3y^2 } \\
\int\limits_0^1 {3y^2 e^{y^3 } - 3y^2 dy} = \left. {e^{y^3 } - y^3 } \right|_0^1 = e - 1 - 1 = e - 2 \\
\end{array}\)

 

[mindy88]

oh i forgot the -3y^2, i thought it was 0

thanks