The area of the region bound by the curves $y=x^{2}$ and $y=4 x-x^{2}$ is

• A

  $\frac{16}{3}$ sq. units
• Correct answer
  B

  $\frac{8}{3}$ sq. units
• C

  $\frac{4}{3}$ sq. units
• D

  $\frac{2}{3}$ sq. unit

Solution

Given equation of curves are $y=x^{2}$ and $\begin{aligned} y &=4 x-x^{2} \\ y-4 &=-4+4 x-x^{2} \\ y-4 &=-\left(x^{2}-4 x+4\right)=-(x-2)^{2} \end{aligned}$

Now, required area $=\int_{0}^{2}\left\{\left(4 x-x^{2}\right)-x^{2}\right\} d x$ $$ \begin{aligned} &=\int_{0}^{2}\left\{\left(4 x-2 x^{2}\right) d x\right. \\ &=\left[2 x^{2}-\frac{2}{3} x^{3}\right]_{0}^{2} \\ &=\left[\left(8-\frac{16}{3}\right)-0\right] \\ &=\frac{24-16}{3}=\frac{8}{3} \text { sq. units } \end{aligned} $$

Hence, the correct answer is option: 'B'.