question_answer

The area of the figure bounded by \[{{y}^{2}}=2x+1\]and \[x-y=1\] is

A) \[\frac{2}{3}\]

B) \[\frac{4}{3}\]

C) \[\frac{8}{3}\]

D) \[\frac{16}{3}\]

Correct Answer: D

Solution :

[d] Area of the region is given by \[A=\int\limits_{-1}^{3}{\left[ (y+1)-\left( \frac{{{y}^{2}}-1}{2} \right) \right]dy=\frac{16}{3}}\]