Integrate $y^2$ over the region in the following image:https://i.stack.imgur.com/CSPyo.png
I'd post it directly but I need ten reputation. I know the answer must be 1/3 geometrically, but the integral I created to solve the problem evaluates to 4/3.
$\int_0^2 \int_{x-1}^x y^2 dy dx$
What am I doing wrong?
On
The correct domain of integral is like \begin{align*} \int_{0}^{1}\int_{0}^{x}y^{2}dydx+\int_{1}^{2}\int_{x-1}^{1}y^{2}dydx. \end{align*}
On
As other answers have mentioned, if you’re going to integrate directly, you need to split the region up so that you have the correct bounds in the inner integral. Or, you can make the substitution $x\to u+v$, $y\to v$, which changes the region into the unit square. The Jacobian determinant of this transformation is $1$, so the integral becomes simply $$\int_0^1\int_0^1v^2\,dv\,du.$$
The upper and lower limit is not always $x$ and $x-1$ respectively.
When $x$ is from $0$ to $1$, it is bounded above by $x$ and bounded below by $0$.
When $x$ is from $1$ to $2$, it is bounded above by $1$ and bounded below by $x-1$.