A) \[1/3\]
B) \[2/3\]
C) \[1\]
D) \[2\]
Correct Answer: B
Solution :
Given curve is \[2x={{y}^{2}}-1\] Required area \[=|\int_{-1}^{1}{x\,dy|}\] \[=\left| \int_{-1}^{1}{\left( \frac{{{y}^{2}}-1}{2} \right)dy} \right|\] \[=\left| 2\int_{0}^{1}{\left( \frac{{{y}^{2}}-1}{2} \right)dy} \right|\] \[=\left| \left[ \frac{{{y}^{3}}}{3}-y \right]_{0}^{1} \right|\] \[=\left| \frac{1}{3}-1 \right|=\frac{2}{3}\,sq\,unit\]
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