I've been struggling for a bit on this math problem from my homework.
$$ \int_{0}^{1}\int_{1}^{3}(8x^3-36x^2y^2)dydx $$
From what I understand and have attempted, we'll integrate with respect to $x$ first, so something like:
$$ (2x^4-12x^3y^2)\vert_{1}^{3}=160-312y^2 $$
and then another integration, this time with respect to $y$:
$$ 160y-104y^3\vert_{0}^{1}=(160-104)-(0-0) = \textbf{56} $$
Yet my book says its wrong.
I could be missing something for all I know, or could it just be something simple I'm overlooking? Regardless, I'm a bit stumped and would appreciate a nudge in the right direction.
Thank you!
The order of integration is from the inside out, so $dy \, dx$ means you integrate with respect to $y$ first, then with respect to $x$.
So the first antiderivative is $$\int 8x^3 - 36 x^2 y^2 \, dy = 8x^3 y - 12 x^2 y^3 + C$$ and you evaluate these at the endpoints $y = 3$ and $y = 1$: $$\int_{y=1}^3 8x^3 - 36 x^2 y^2 \, dy = \left[8x^3 y - 12 x^2 y^3 \right]_{y=1}^3 = 16x^3 - 312 x^2.$$ The remainder of the computation I leave as an exercise.