question_answer

Area bounded by the curve \[{{x}^{4}}=4y\]and the straight line \[x=4y-2\]is:

A) \[\frac{\text{8}}{\text{9}}\,\text{sq}\,\text{unit}\]                         

B)  \[\frac{9}{8}\,\text{sq}\,\text{unit}\]   

C)         \[\frac{4}{3}\,\text{sq}\,\text{unit}\]   

D)         none of these

Correct Answer: B

Solution :

The point of intersection of the curves \[{{x}^{2}}=4y\] and \[x=4y-2\]is (2, 1) and \[\left( -,\frac{1}{4} \right)\] \[\therefore \]Required area \[=\int_{-1}^{2}{({{y}_{2}}-{{y}_{1}})dx}\] \[=\int_{1}^{2}{\left[ \left( \frac{x+2}{4} \right)-\left( \frac{{{x}^{2}}}{4} \right) \right]}\,dx\] \[=\int_{-1}^{2}{\left( \frac{x}{4}+\frac{1}{2}-\frac{{{x}^{2}}}{4} \right)}\,dx\] \[=\left[ \frac{{{x}^{2}}}{8}+\frac{1}{2}x-\frac{{{x}^{3}}}{12} \right]_{-1}^{2}\] \[=\left[ \frac{1}{2}+1-\frac{2}{3}-\left( \frac{1}{8}-\frac{1}{2}+\frac{1}{12} \right) \right]\] \[=\frac{5}{6}+\frac{7}{24}=\frac{20+7}{24}\] \[=\frac{27}{24}=\frac{9}{8}\,\text{sq}\,\text{unit}\]