JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 8)
The area of the region given by $$\{ (x,y):xy \le 8,1 \le y \le {x^2}\} $$ is :
$$16{\log _e}2 - {{14} \over 3}$$
$$8{\log _e}2 - {{13} \over 3}$$
$$16{\log _e}2 + {7 \over 3}$$
$$8{\log _e}2 + {7 \over 6}$$
Explanation
_1st_February_Evening_Shift_en_8_1.png)
$\begin{aligned} & \text { Required area }=\int_1^2\left(x^2-1\right) d x+\int_2^8\left(\frac{8}{x}-1\right) d x \\\\ & =\left.\left(\frac{x^3}{3}-x\right)\right|_1 ^2+\left.(8 \ln x-x)\right|_2 ^8 \\\\ & =\left[\left(\frac{8}{3}-2\right)-\left(\frac{1}{3}-1\right)\right]+[8 \ln 8-8-(8 \ln 2-2)] \\\\ & =\frac{4}{3}+8 \ln 4-6 \\\\ & =8 \ln 4-\frac{14}{3} \\\\ & =16 \ln 2-\frac{14}{3}\end{aligned}$
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