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\[\int_{-1}^{3}{({{x}_{2}}-{{x}_{1}})\,\,dy}\] \[=\int_{-1}^{3}{\left\{ (y+1)-\frac{({{y}^{2}}-1)}{2} \right\}dy}\] \[=\frac{1}{2}\int_{-1}^{3}{(2y+3-{{y}^{2}})\,\,dy}\] \[=\frac{1}{2}\left[ {{y}^{2}}+3y-\frac{{{y}^{3}}}{3} \right]_{-1}^{3}\] \[=\frac{1}{2}\left[ (9+9-9)-\left( 1-3+\frac{1}{3} \right) \right]\] \[=\frac{1}{2}\left( 9+\frac{5}{3} \right)\] \[=\frac{16}{3}sq\,\,units\]You need to login to perform this action.
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