A) \[\frac{14}{3}\] sq. unit
B) \[\frac{3}{4}\] sq. unit
C) \[\frac{3}{16}\] sq. unit
D) \[\frac{16}{3}\] sq. unit
Correct Answer: D
Solution :
Equations of curves \[{{y}^{2}}=4x\] and \[{{x}^{2}}=4y.\]The given equations may be written as \[y=2\sqrt{x}\] and \[y=\frac{{{x}^{2}}}{4}.\] We know that area enclosed by the parabolas \[=\int_{\,0}^{\,4}{2\sqrt{x}\,dx-}\int_{\,0}^{\,4}{\frac{{{x}^{2}}}{4}dx=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}}\]sq. unit.
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