Question

If $S$ is the sum of cubes of possible value of $c$ for which the area of the figure bounded by the curve $y=8x^2-x^5$, the straight lines $x=1$ and $x=c$ and the abscissa axis is equal to $\frac{16}{3}$, then the value of $\left[S\right]$, where $[\cdot ]$ denotes the greatest integer funtion, is _______

Answer 1

Area under curve $y=8x^2-x^5$ between $x=1$ and $x=c$ and abscissa axis

$={\int }_{x=1}^{x=c}ydx={\int }_{x=1}^{x=c}(8x^2-x^5)dx={[8\frac{x^3}{3}-\frac{x^6}{6}]}_{x=1}^{x=c}=[\frac{8c^3}{3}-\frac{c^6}{6}]-[\frac{8}{3}-\frac{1}{6}]=\frac{8c^3}{3}-\frac{c^6}{6}-\frac{15}{6}$

According to question, Area under curve = $\frac{16}{3}$

$\Rightarrow \frac{8c^3}{3}-\frac{c^6}{6}-\frac{15}{6}=\frac{16}{3}$

$\Rightarrow \frac{8c^3}{3}-\frac{c^6}{6}=\frac{16}{3}+\frac{15}{6}$

$\Rightarrow \frac{8c^3}{3}-\frac{c^6}{6}=\frac{47}{6}$

$\Rightarrow c^3(8-\frac{c^3}{2})=\frac{47}{2}$

Let $c^3$ be $z$

$\therefore z(8-\frac{z}{2})=\frac{47}{2}\Rightarrow z(16-z)=47\Rightarrow z^2-16z+47$

Now, Sum of roots of above quadratic equation $=\frac{-b}{a}=-\frac{(-16)}{1}=16$

$\because z$ is $c^3$

$\therefore$ Sum of possible values of $z$ is the sum of cubes of possible values of $c$, which according to the question is equal to $S$

Hence, $S=16$ and $\left[S\right]=\left[16\right]=16$.