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)         \[\frac{11}{3}\]

Correct Answer: D

Solution :

Option is not correct Given curves are \[{{y}^{2}}=2x+1\] and        \[x-y=1\] Points of intersection are \[A(0,-1)\]and \[B(4,3).\] Area \[=\int_{-1}^{3}{(1+y)dy-\int_{-1}^{3}{\left( \frac{{{y}^{2}}-1}{2} \right)dy}}\]                 \[=\left[ y+\frac{{{y}^{2}}}{2} \right]_{-1}^{3}-\left[ \frac{1}{2}\left( \frac{{{y}^{3}}}{3}-y \right) \right]_{-1}^{3}\] \[=\left[ 3+\frac{9}{2}-\left( -1\frac{1}{2} \right) \right]-\frac{1}{2}\left[ 9-3-\left( -\frac{1}{3}+1 \right) \right]\]                 \[=8-\frac{8}{3}=\frac{16}{3}\]