Question

The 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 $x-$axis is equal to $\frac{16}{3}$ is

• A. $2$
• B. $\sqrt{8-\sqrt{17}}$
• C. $3$
• D. $-1$

Answer 1

The correct option is D $-1$

For $c<1,{\int }_c^1(8x^2-x^5)dx=\frac{16}{3}$(given)
$\Rightarrow {[\frac{8x^3}{3}-\frac{x^6}{6}]}_c^1$
$\Rightarrow \frac{8}{3}-\frac{1}{6}-\frac{8c^3}{3}+\frac{c^6}{6}=\frac{16}{3}$
$\Rightarrow c^3[-\frac{8}{3}+\frac{c^3}{6}]=\frac{16}{3}-\frac{8}{3}+\frac{1}{6}=\frac{17}{6}$
$\Rightarrow -\frac{8}{3}{c}^3+\frac{c^6}{6}=\frac{17}{6}$
$\Rightarrow c^6-16c^3-17=0$
On factorization, we get $(c^3+1)(c^3-17)=0$
$\Rightarrow c^3=-1,17$
$\Rightarrow c=-1,{17}^{\frac{1}{3}}$
$\Rightarrow c=-1$ satisfy the above equation.
For $c\ge 1,$ none of the values of $c$ satisfies the required condition that
${\int }_1^c(8x^2-x^5)dx=\frac{16}{3}$
$\therefore c=-1$