JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 8)
The area of the region given by
$$A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right\}$$ is :
$$\frac{31}{8}$$
$$\frac{17}{6}$$
$$\frac{19}{6}$$
$$\frac{27}{8}$$
Explanation
$A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right.$
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So area of required region
$$ \begin{aligned} &A=\int_{-1}^{\frac{1}{2}}\left(x+2-x^{2}\right) d x+\int_{\frac{1}{2}}^{1}\left(4-3 x-x^{2}\right) d x \\\\ &=\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{\frac{1}{2}}+\left[4 x-\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{\frac{1}{2}}^{1} \\\\ &=\left(\frac{1}{8}+1-\frac{1}{24}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+\left(4-\frac{3}{2}-\frac{1}{3}\right)-\left(2-\frac{3}{8}-\frac{1}{24}\right) \\\\ &=\frac{17}{6} \end{aligned} $$
So area of required region
$$ \begin{aligned} &A=\int_{-1}^{\frac{1}{2}}\left(x+2-x^{2}\right) d x+\int_{\frac{1}{2}}^{1}\left(4-3 x-x^{2}\right) d x \\\\ &=\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{\frac{1}{2}}+\left[4 x-\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{\frac{1}{2}}^{1} \\\\ &=\left(\frac{1}{8}+1-\frac{1}{24}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+\left(4-\frac{3}{2}-\frac{1}{3}\right)-\left(2-\frac{3}{8}-\frac{1}{24}\right) \\\\ &=\frac{17}{6} \end{aligned} $$
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