question_answer

The area (in sq. units) of the region \[A=\{(x,y):\frac{{{y}^{2}}}{2}\le x\le y+4\}\]is :-             [JEE Main 9-4-2019 Afternoon]

A) \[\frac{53}{3}\]                                  

B) 18

C) 30                               

D) 16

Correct Answer: B

Solution :

\[{{y}^{2}}=2x\] \[x-y-4=0\] \[{{(x-4)}^{2}}=2x\] \[{{x}^{2}}+16-8x-2x=0\] \[{{x}^{2}}-10x+16=0\] \[x=8,2\] \[y=4,-2\] \[A=\int\limits_{-2}^{4}{\left( y+4-\frac{{{y}^{2}}}{2} \right)dy}\] \[\left. =\frac{{{y}^{2}}}{2} \right|_{-2}^{4}+4y|_{-2}^{4}-\left. \frac{{{y}^{3}}}{6} \right|_{-2}^{4}\] \[=(8-2)+4(6)-\frac{1}{6}(64+8)\] \[=6+24-12=18\]