Area Under Curves Question 4
Question 4 - 2024 (27 Jan Shift 1)
Let the area of the region ${(x, y): x-2 y+4 \geq 0, x+2 y^{2} \geq 0, x+4 y^{2} \leq 8, y \geq 0 }$ be $\frac{m}{n}$, where $m$ and $n$ are coprime numbers. Then $m+n$ is equal to
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Answer (119)
Solution
$$ \begin{aligned} & A=\int _0^{1}\left[\left(8-4 y^{2}\right)-\left(-2 y^{2}\right)\right] d y+ \\ & \int _1^{3 / 2}\left[\left(8-4 y^{2}\right)-(2 y-4)\right] d y \\ & =\left[8 y-\frac{2 y^{3}}{3}\right] _0^{1}+\left[12 y-y^{2}-\frac{4 y^{3}}{3}\right] _1^{3 / 2}=\frac{107}{12}=\frac{m}{n} \\ & \therefore m+n=119 \end{aligned} $$
