Trouble with double integration

Question

I'm simply trying to compute the following double integral:

$$ \int_1^4\int_0^3\ (\ x\ +\ 2y\ )\ dx\ dy $$

And here are my steps:

$$ \int_1^4\ \left.(\ \frac{1}{2}x^2\ +\ 2xy\ )\right|_0^3\ dy $$ $$ \int_1^4\ (\ \frac{9}{2}\ +\ 6y\ )\ dy $$ $$ \frac{9}{2}\ +\ 6\int_1^4\ y\ dy $$ $$ \frac{9}{2}\ +\ 6\ \frac{1}{2}\left.(\ y^2\ )\right|_1^4 $$ $$ \frac{9}{2}\ +\ 3(\ 16\ -\ 1\ ) $$ $$ \frac{9}{2}\ +\ 3(\ 15\ ) $$ $$ \frac{9}{2}\ +\ 45 $$ $$ \frac{9\ + 90}{2} $$ $$ \frac{99}{2} $$

The answer according to my book is 117 / 2, however.

$$ \int_1^4\int_0^3\ (\ x\ +\ 2y\ )\ dx\ dy\ =\ \frac{117}{2} $$

What am I doing wrong?

Answer 1

You go from $$ \int_1^4\ (\ \frac{9}{2}\ +\ 6y\ )\ dy $$ to $$ \frac{9}{2}\ +\ 6\int_1^4\ y\ dy $$ which is incorrect. The second step should just be evaluating the antiderivative of $\frac{9}{2} + 6y$ as $$ (\frac{9}{2}y + 3\left. y^2) \right|_1^4 $$ Your work seemed fine otherwise, so I assume you can finish it with that correction.