question_answer

The area (in sq. units) of the region bounded by the curve \[{{x}^{2}}=4y\]and the straight line\[x=4y-2\]is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

A) \[\frac{9}{8}\]                                      

B)               \[\frac{3}{4}\]

C)               \[\frac{5}{4}\]                                      

D)               \[\frac{7}{8}\]

Correct Answer: A

Solution :

Given equation of parabola is \[{{x}^{2}}=4y\]and equation of straight line is \[x=4y-2.\] \[\therefore \]\[{{x}^{2}}=x+2\] \[\Rightarrow \]\[{{x}^{2}}-x-2=0\] \[\Rightarrow \]\[x=2,-1\]       So, required area \[=\int\limits_{-1}^{2}{\left( \frac{x+2}{4}-\frac{{{x}^{2}}}{4} \right)dx}\] \[=\frac{1}{4}\int\limits_{-1}^{2}{\left( x+2-{{x}^{2}} \right)dx}\] \[=\frac{1}{4}\left[ \frac{{{x}^{2}}}{2}+2x-\frac{{{x}^{3}}}{3} \right]_{-1}^{2}=\frac{9}{8}sq.\]units