Question

The area of the region bounded by the curves $x^2=2y-x$ and x-axis (in square units), is

• A. $\frac{4}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2\pi }{3}-\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{3}$

$x^2=2y-x\Rightarrow x^2+x=2y$

At x axis ( $y=0$ )

$x^2=-x\Rightarrow x=0,-1$

$\therefore$ The area $={\int }_{-1}^{0}y.dx={\int }_{-1}^0\frac{1}{2}(x^2+x)dx=\frac{1}{2}.{[\frac{x^3}{3}+\frac{x^2}{2}]}_{-1}^0=\frac{1}{2}\left[\right(0+0)-(\frac{-1}{3}+\frac{1}{2})]=\frac{1}{2}.(\frac{1}{3}-\frac{1}{2})=\frac{1}{2}.\frac{-1}{6}=\frac{-1}{12}$

SInce area is positive so. area $=\left|\frac{-1}{12}\right|=\frac{1}{12}sq.units.$