TITLE: Integral of the level curve $ye^{2x}$
QUESTION [0 upvotes]: $\int_{0}^{2}\int_{0}^{4} ye^{2x}$ $dydx$
The book says the answer is $32(e^4-1)$
I had to do u-substitution when doing it, maybe that's where I went wrong.
First integrating with respect to y, holding $e^{2x}$ as a constant and plugging in gives me:
$\int_0^2\int_0^4 \frac{1}{2}y^2e^{2x} \Big\vert_0^4 \to \int_0^2 8e^{2x}dx$
This quit clearly calls for u substitution here, from what I remember 2x is a function inside of $f(x)=e^x$ where $g(x)=2x$ and $f(g(x))=e^{2x}$:
$u=2x$
$du=2dx \to \frac{1}{2}du=dx$
Getting my bounds: $u(2)=4$ and $u(0)=0$
My new integral is:
$\int_0^4 \frac{1}{2} (8)e^u$ $\to$ $\int_0^4 4e^u \Big\vert_0^4=4e^8-4=4(e^8-1)$

REPLY [2 votes]: The book is wrong, your derivation is also wrong.
Note that you substitute $u=8$ instead of $u=4$. You already changed the bounds to integrate from $u=0$ to $4$. The correct answer is $4(e^4 - 1)$.
$\int_0^4 \frac{1}{2} (8)e^u du$ $\to$ $\int_0^4 4e^u du \Big\vert_0^4=4e^\color{red}{4}-4=4(e^\color{red}{4}-1)$

REPLY [0 votes]: Slight error in your final line. The original x limits were 0 and 2 which changed to 0 and 4 for u. However you seemed to have doubled them again when substituting into $4e^u$.
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