question_answer

The area bounded by the curves \[y=\sqrt{|x|}\] and \[y=\pm x,\] is

A)  \[0\]

B)  \[\frac{1}{6}\]

C)  \[\frac{1}{3}\]

D)  \[\frac{2}{3}\]

Correct Answer: B

Solution :

Given,  curves are \[y=\sqrt{|x|}\] \[\Rightarrow \] \[{{y}^{2}}=x,\] which is a parabola and \[y=\pm x\] Intersecting point of parabola and line is \[A(1,\,1).\] \[\therefore \]  Required area = (Area under parabola) -(Area under line) \[=\int_{0}^{1}{(\sqrt{x})\,dx-\int_{0}^{1}{x\,\,dx=\int_{0}^{1}{{{x}^{1/2}}\,dx-\int_{0}^{1}{x\,\,dx}}}}\] \[=\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{1}-\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}=\frac{2}{3}[{{x}^{3/2}}]_{0}^{1}-\frac{1}{2}[{{x}^{2}}]_{0}^{1}\] \[=\frac{2}{3}[{{(1)}^{3/2}}-{{(0)}^{3/2}}]-\frac{1}{2}[{{(1)}^{2}}-{{(0)}^{2}}]\] \[=\frac{2}{3}[1-0]-\frac{1}{2}[1-0]=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}sq\,unit\]