MCQ
Area bounded by parabola ${y^2} = x$ and straight line $2y = x$ is
- ✓$\frac{4}{3}$
- B$1$
- C$\frac{2}{3}$
- D$\frac{1}{3}$
$\Rightarrow {y^2} = 2y \Rightarrow y = 0,\,2$
$\therefore \,$ Required area $ = \int_0^2 {({y^2} - 2y)dy = \left( {\frac{{{y^3}}}{3} - {y^2}} \right)_0^2}$
$={ \frac{4}{3}} \,\, sq. \,unit$.
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$\sqrt {5\,x\,\, - \,\,6\,\, - \,\,{x^2}} \,\, + \,\,\frac{\pi }{2}\,\,\int\limits_0^x {} $$dz > x \int\limits_0^\pi {} sin^2 x \,dx$ is :