Verified Solution • 2025–26 Session moderate

QJEEM22Ma3427: JEE Main 2022 Mathematics - Class 12 JEEM22Ma3427 Solution: The area of the region given by is :

The area of the region given by

A={(x,y):x2≤y≤min⁡{x+2,4−3x}}A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right\} is :

Multiple Choice Options

  • A 318\frac{31}{8}
  • B 176\frac{17}{6}
  • C 196\frac{19}{6}
  • D 278\frac{27}{8}

Step-by-Step Verified Solution

A={(x,y):x2≤y≤min⁡{x+2,4−3x}A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right.

JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Area Under The Curves Question 68 English Explanation

So area of required region

amp;A=∫−112(x+2−x2)dx+∫121(4−3x−x2)dxamp;=[x22+2x−x33]−112+[4x−3x22−x33]121amp;=(18+1−124)−(12−2+13)+(4−32−13)−(2−38−124)amp;=176\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}

Related Learning Resources & Exam Preparation