A) 16/3 sq unit
B) 32/3 sq unit
C) 8/3 sq unit
D) 64/3 sq unit
E) 4/3 sq unit
Correct Answer: B
Solution :
The given equation of parabola is \[{{y}^{2}}=8x\] ...(i) Required area \[=2\int_{0}^{2}{\sqrt{8x}}dx=2.2\sqrt{2}\int_{0}^{2}{\sqrt{x}}dx\] \[=4\sqrt{2}\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{2}=4\sqrt{2}\left[ \frac{2\sqrt{2}}{3/2} \right]\] \[=\frac{32}{3}sq\,unit\]You need to login to perform this action.
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