Q. The area of the region bounded by the curves $y = x^2$ and $y = 4x - x^2$ in square units is
Solution:
$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 $ = \int\limits_0^2 (4x - x^2) dx - \int\limits_0^2 x^2 dx$
$= \left| \frac{4x^{2}}{2} - \frac{x^{3}}{3}\right|^{2}_{0} - \left|\frac{x^{3}}{3}\right|^{2}_{0} $
$=2.\left(4\right) - \frac{8}{3}-\frac{1}{3} \left(8\right) =8 - \frac{16}{3}= \frac{8}{3}$
