question_answer

    Area bounded by the curve \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] and the straight line \[0<x<\frac{\pi }{8}\], is equal to :

A)  \[\frac{\pi }{4}<x<\frac{3\pi }{8}\] sq. units     

B)  \[\frac{3\pi }{8}<x<\frac{5\pi }{8}\]sq. units

C)  \[\frac{5\pi }{8}<x<\frac{3\pi }{4}\]sq. units     

D)  none of these

Correct Answer: B

Solution :

                Solving the equations\[{{x}^{2}}=4y\] and\[x=4y-2,\]then the point of intersection of the parabola are the line are \[A(2,1)\]and\[B\left( -1,\frac{1}{4} \right)\] \[\therefore \]The required area = shaded area \[=\left[ \int_{-1}^{2}{ydx} \right]-\left[ \int_{-1}^{2}{ydx}(from\,{{x}^{2}}=4y) \right]\] \[=\int_{-1}^{2}{\frac{1}{4}}(x+2)dx-\int_{-1}^{2}{\frac{1}{4}}{{x}^{2}}dx\] \[=\frac{1}{4}\left[ \frac{{{x}^{2}}}{2}+2x \right]_{-1}^{2}-\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{-1}^{2}\] \[=\frac{9}{8}\]sq. units