A) \[\frac{8}{9}\] sq. unit
B) \[\frac{9}{8}\] sq. unit
C) \[\frac{4}{3}\] sq. unit
D) None of these
Correct Answer: B
Solution :
Solving the equations \[{{x}^{2}}=4y\]and \[x=4y-2\] simultaneously. The points of intersection of the parabola and the line are \[A(2,\,1)\]and \[B\left( -1,\frac{1}{4} \right)\]. \ The required area = shaded area \[=\left[ \int_{-1}^{2}{y\,d{{x}_{(\text{for}\,x=4y-2)}}} \right]-\left[ \int_{-1}^{2}{yd{{x}_{(\text{for}\,{{x}^{2}}=4y)}}} \right]\] \[=\int_{-1}^{2}{\frac{1}{4}(x+2)dx-\int_{-1}^{2}{\frac{1}{4}{{x}^{2}}dx}}\] \[=\frac{1}{4}\left[ \frac{{{x}^{2}}}{2}+2x \right]_{-1}^{2}-\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{-1}^{2}=\frac{9}{8}\]sq. unit.You need to login to perform this action.
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