The value of $$c + 2$$  for which the area of the figure bounded by the curve $$y = 8{x^2} - {x^5},$$   the straight lines $$x = 1$$  and $$x = c$$  and $$x$$-axis is equal to $$\frac{{16}}{3},$$ is :

Solution :
$$\eqalign{ & {\text{For }}c < 1,\,\int_c^1 {\left( {8{x^2} - {x^5}} \right)} dx = \frac{{16}}{3} \cr & \Rightarrow \frac{8}{3} - \frac{1}{6} - \frac{{8{c^3}}}{3} + \frac{{{c^6}}}{6} = \frac{{16}}{3} \cr & \Rightarrow {c^3}\left[ { - \frac{8}{3} + \frac{{{c^3}}}{6}} \right] = \frac{{17}}{6} \cr} $$
Again, for $$c \geqslant 1,$$  none of the values of $$c$$ satisfy the required condition that
$$\int_1^c {\left( {8{x^2} - {x^5}} \right)} dx = \frac{{16}}{3} \Rightarrow c + 2 = 1$$