Area Under Curves Question 4
Question 4 - 2024 (27 Jan Shift 1)
Let the area of the region $\left{(x, y): x-2 y+4 \geq 0, x+2 y^{2} \geq 0, x+4 y^{2} \leq 8, y \geq 0\right}$ be $\frac{m}{n}$, where $m$ and $\mathrm{n}$ are coprime numbers. Then $\mathrm{m}+\mathrm{n}$ is equal to
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Answer (119)
Solution
$$ \begin{aligned} & A=\int_{0}^{1}\left[\left(8-4 y^{2}\right)-\left(-2 y^{2}\right)\right] d y+ \ & \int_{1}^{3 / 2}\left[\left(8-4 y^{2}\right)-(2 y-4)\right] d y \ & =\left[8 y-\frac{2 y^{3}}{3}\right]{0}^{1}+\left[12 y-y^{2}-\frac{4 y^{3}}{3}\right]{1}^{3 / 2}=\frac{107}{12}=\frac{m}{n} \ & \therefore m+n=119 \end{aligned} $$
