JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 11)
The area of the region $$\left\{(x, y): x^{2} \leq y \leq\left|x^{2}-4\right|, y \geq 1\right\}$$ is
$$\frac{4}{3}(4 \sqrt{2}+1)$$
$$\frac{3}{4}(4 \sqrt{2}+1)$$
$$\frac{4}{3}(4 \sqrt{2}-1)$$
$$\frac{3}{4}(4 \sqrt{2}-1)$$
Explanation
_13th_April_Evening_Shift_en_11_1.png)
$$ \begin{aligned} & \text { Required area }=2\left[\int_1^2 \sqrt{y} d y+\int_4^2 \sqrt{4-y} d y\right] \\\\ & \left.\left.=2\left[\frac{y^{3 / 2}}{\frac{3}{2}}\right]_1^2-\frac{2(4-y)^{3 / 2}}{3}\right]_2^4\right] \\\\ & =\frac{4}{3}(4 \sqrt{2}-1) \end{aligned} $$
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