Question

The area of the region bounded by the curve $y=2x-x^2$ and x-axis is

• A. $\frac{2}{3}$ sq. units
• B. $\frac{4}{3}$ sq. units
• C. $\frac{5}{3}$ sq. units
• D. $\frac{8}{3}$ sq. units

Answer 1

Answer: (B)

$\frac{4}{3}$ sq. units

To get intersection with x-axis, put $y=0$

$2x-x^2=0$$⟹x(2-x)=0$

$x=0$ and $2$

Now making the graph we see that the curve makes an intercept between $(0,0)$ and $(2,0)$.

So area

$={\int }_0^2(2x-x^2).dx$

$={[x^2-x^3/3]}_0^2$

$=4-\frac{8}{3}-0$

$=\frac{4}{3}$

answer (b)