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QJEEM23Maf01e: JEE Main 2023 Mathematics - Class 12 JEEM23Maf01e Solution: The area of the region given by is :

The area of the region given by {(x,y):xy≤8,1≤y≤x2}\{ (x,y):xy \le 8,1 \le y \le {x^2}\} is :

Multiple Choice Options

  • A 16log⁡e2−14316{\log _e}2 - {{14} \over 3}
  • B 8log⁡e2−1338{\log _e}2 - {{13} \over 3}
  • C 16log⁡e2+7316{\log _e}2 + {7 \over 3}
  • D 8log⁡e2+768{\log _e}2 + {7 \over 6}

Step-by-Step Verified Solution

JEE Main 2023 (Online) 1st February Evening Shift Mathematics - Area Under The Curves Question 58 English Explanation

amp; Required area =∫12(x2−1)dx+∫28(8x−1)dxamp;=(x33−x)∣12+(8ln⁡x−x)∣28amp;=[(83−2)−(13−1)]+[8ln⁡8−8−(8ln⁡2−2)]amp;=43+8ln⁡4−6amp;=8ln⁡4−143amp;=16ln⁡2−143\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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