Jee Main 2024 29 01 2024 Shift 2 - Question12
Question 12
Area bounded by $0 \leq y \leq \min {x^{2}+2,2 x+2 }, x \in$ $[0,3]$ is $A$, then $12 A$ is
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Answer: (164)
Solution:
$\min {x^{2}+2,2 x+2 } \begin{cases}x^{2}+2 & 0 \leq x \leq 2 \\ 2 x+2 & 2 \leq x \leq 3\end{cases}$
Area $=A=\int _0^{2}\left(x^{2}+2\right) d x+\frac{1}{2}[6+8] \times 1$
$\left.=\frac{x^{3}}{3}+2 x\right] _0^{2}+7$ $\frac{8}{3}+4+7=\left(\frac{8}{3}+11\right)$ unit
$12 A=12\left(\frac{8}{3}+11\right)=164$
BETA
