Question

The area of the region bounded by the curves $y=x^2$ and $y=4x-x^2$ in square units is

• A. $\frac{1}{3}$
• B. $\frac{16}{3}$
• C. $\frac{8}{3}$
• D. $\frac{4}{3}$

Answer 1

Answer: (C)

$y=x^2$ ...(i)
$y=4x-x^2$ ....(ii)
Their intersection point can be given by, $x^2=4x-x^2\Rightarrow 2x^2-4x=0$
i.e., two intersection points are (0, 0) and (2, 4)
$\therefore$ Required area $=\underset{0}{\overset{2}{\int }}(4x-x^2)dx-\underset{0}{\overset{2}{\int }}{x}^{2}dx$
$={\left|\frac{4x^2}{2}-\frac{x^3}{3}\right|}_0^2-{\left|\frac{x^3}{3}\right|}_0^2$
$=2.\left(4\right)-\frac{8}{3}-\frac{1}{3}\left(8\right)=8-\frac{16}{3}=\frac{8}{3}$