question_answer

The area bounded by the curve\[{{y}^{2}}=2x+1\]and the straight line\[x-y-1=0\]is given by

A) \[\frac{9}{2}sq\,\,units\]                             

B) \[\frac{43}{6}sq\,\,units\]

C) \[\frac{35}{6}sq\,\,units\]           

D)         None of these

Correct Answer: D

Solution :

The point of intersection of curve\[{{y}^{2}}=2x+1\]and line\[x-y-1=0\]is (14, 3) and (0, -1) \[\therefore \]Required area\[\int_{-1}^{3}{({{x}_{2}}-{{x}_{1}})\,\,dy}\]                 \[=\int_{-1}^{3}{\left\{ (y+1)-\frac{({{y}^{2}}-1)}{2} \right\}dy}\]                 \[=\frac{1}{2}\int_{-1}^{3}{(2y+3-{{y}^{2}})\,\,dy}\]                 \[=\frac{1}{2}\left[ {{y}^{2}}+3y-\frac{{{y}^{3}}}{3} \right]_{-1}^{3}\]                 \[=\frac{1}{2}\left[ (9+9-9)-\left( 1-3+\frac{1}{3} \right) \right]\]                 \[=\frac{1}{2}\left( 9+\frac{5}{3} \right)\]                 \[=\frac{16}{3}sq\,\,units\]