Hello, duude!
$\displaystyle \int^1_0\int^x_{x^2}\frac{x^3y}{1y^2}\,dy\,dx$
By changing the order of integration, show that the value of the integral is $\displaystyle \frac{1}{16}$ The limits are: .$\displaystyle \begin{Bmatrix}x = 1 \\ x=0\end{Bmatrix} \text{ and }\begin{Bmatrix}y = x \\ y = x^2\end{Bmatrix}$
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Reversing the limits, we have: .$\displaystyle \begin{Bmatrix}y = 1 \\ y = 0\end{Bmatrix}\text{ and }\begin{Bmatrix}x = \sqrt{y} \\ x = y\end{Bmatrix}$
And the integral is: .$\displaystyle \int^1_0\int^{\sqrt{y}}_y\frac{x^3y}{1y^2}\,dx\,dy$
The inner integral is: .$\displaystyle \int^{\sqrt{y}}_y\frac{x^3y}{1y^2}\,dx \;\;=\;\;\frac{1}{4}\cdot\frac{x^4y}{1y^2}\,\bigg^{\sqrt{y}}_y \;\;=\; \;\frac{1}{4}\,\left[\frac{(\sqrt{y})^4y}{1y^2}  \frac{y^4\cdot y}{1y^2}\right] $
. . $\displaystyle = \;\;\frac{1}{4}\left[\frac{y^3}{1y^2}  \frac{y^5}{1y^2}\right] \;\;=\;\;\frac{1}{4}\left[\frac{y^3(1y^2)}{1y^2}\right] \;\;=\;\;\frac{1}{4}y^3 $
Then we have: .$\displaystyle \int^1_0\frac{1}{4}y^3\,dy \;\;=\;\;\frac{1}{16}y^4\,\bigg^1_0 \;\;=\;\;\frac{1}{16}$