Q102 / 180 · KCET 2008 · Mathematics
The area of the region bounded by $y=2 x-x^{2}$ and the $x$-axis is
- A$\frac{8}{3}$ sq unit
- Correct answerB$\frac{4}{3}$ sq unit
- C$\frac{7}{3}$ sq unit
- D$\frac{2}{3}$ sq unit
Solution
Given curve is
$$
y=2 x-x^{2}
$$
or
$$
(x-1)^{2}=-(y-1)
$$
The curve cut the $x$-axis at $(0,0)$ and $(2,0)$. $\therefore$ Required area $=\int_{0}^{2} y \mathrm{dx}$ $$ \begin{aligned} &=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} \\ &=4-\frac{8}{3}=\frac{4}{3} \text { sq unit } \end{aligned} $$
The curve cut the $x$-axis at $(0,0)$ and $(2,0)$. $\therefore$ Required area $=\int_{0}^{2} y \mathrm{dx}$ $$ \begin{aligned} &=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} \\ &=4-\frac{8}{3}=\frac{4}{3} \text { sq unit } \end{aligned} $$
Hence, the correct answer is option: 'B'.