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Solve the area of the shaded region.

Solution

VerifiedAs we can see the points of intersection are $(-3, 3)$ and $(0, 0)$. The area of the region sought is best to evaluate by integrating with respect to $y$ since the equations are expressed as $x$ in terms of $y$. We also can rewrite them $x_R = 2y-y^2$ (because this curve is on the right) and $x_L = y^2 - 4y$ (because this curve is on the left). So a typical rectangle will have a length of $x_R - x_L = 2y-y^2 - (y^2 - 4y) = 6y - 2y^2$ and width of $\Delta y$ (or $dy$ in integration). Therefore, the area of the region bounded by the two curves is

$\begin{align*} A &=\int_{0}^{3} \left(6y - 2y^2\right) dy\\ &=\left[ 3y^2 - \dfrac{2y^3}{3}\right]_{0}^{3}\\ &=\left(3(3)^3 - \dfrac{2(3)^3}{3} \right) -\left(3(0)^3 - \dfrac{2(0)^3}{3} \right)\\ &=9 - 0\\ &=9 \end{align*}$

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