QJEEM21Mab107: JEE Main 2021 Mathematics - Class 12 JEEM21Mab107 Solution: Let f : [ 3, 1] R be given as f ( x ) = { min { ( x + 6 ) , x 2 …

Step-by-Step Verified Solution

Area is $\int\limits_{ - 3}^{ - 2} {(x + 6)dx + \int\limits_{ - 2}^0 {{x^2}dx + \int\limits_0^1 {\sqrt {x}dx = A} } }$

$= {7 \over 2} + \left[ {{{{x^3}} \over 3}} \right]_{ - 2}^0 + \left[ {{2 \over 3}{x^{3/2}}} \right]_0^1$

$= {7 \over 2} + {8 \over 3} + {2 \over 3} = {{41} \over 6}$

So, 6A = 41