I want to calculate this integral.
$$ \int\int_B x^2y \space dx dy$$ where $$B = \{ (x,y) \in \mathbb{R}^2 | y \leq x \leq y^2+1, 0 \leq y \leq 1 \}$$
Now from the definition we know that the "$y$"-integral needs to be from $0$ to $1$.
I have tried to calculate it like this
$$ \int_0^1 \int_y^{y^2+1}x^2y \ \ dx dy$$
But since I know the answer to be $-\frac{1}{40}$ I knew I was wrong when I got $\frac{127}{120}$ as an answer, so I know I did something wrong with the limits.
Can someone help me ? I have done the calculation on paper, and its too long to type it in here as it is incorrect.
Here are the various steps I did as proof for my work:
$$ \int_0^1 \int_y^{y^2+1}x^2y \ \ dx dy$$
$$ \int_0^1 \left[ \frac{1}{3} x^3y \right]^{x=y^2+1}_{x=y} dy$$
$$ \int_0^1 \left(\frac{1}{3} (y^2+1)^3y \right) - \left(\frac{1}{3}y^4 \right) dy$$
$$ \left[ \frac{1}{24} y^8 + \frac{1}{4}y^4+ \frac{1}{3}y^3 +\frac{1}{2}y^2 -\frac{1}{15}y^5 \right]^1_0 =\frac{127}{120} $$
If $$B = \{ (x,y) \in \mathbb{R}^2 : y \leq x \leq y^2, 0 \leq y \leq 1 \}, $$ then you should be integrating: $$ \int_0^1 \int_{y}^{y^2} x^2y \ dx dy = - \int_0^1 \int_{y^2}^{y} x^2y \ dx dy = -\frac{1}{40}. $$
But if $$B = \{ (x,y) \in \mathbb{R}^2 : y \leq x \leq y^2\color{blue}{+1}, 0 \leq y \leq 1 \}, $$ then you should be integrating: $$ \int_0^1 \int_{y}^{y^2+1} x^2y \ dx dy = \int_0^1 \int_{0}^{x} x^2y \ dy dx + \int_1^2 \int_{\sqrt{x-1}}^{1} x^2y \ dy dx = \frac{67}{120}. $$