Question

Find the value of 'c' for which the area of the figure bounded by the curve, $y-8x^2-x^5$, the straight line $x=1$ & $x=c$ & the abscissa axis is equal to $16/3$

• A. $C=1$ or ${[8-\sqrt{17}]}^{1/3}$
• B. $C=-1$ or ${[8+\sqrt{17}]}^{1/3}$
• C. $C=1$ or ${[8-\sqrt{17}]}^{-1/3}$
• D. $0$

Answer 1

The correct option is A $C=1$ or ${[8-\sqrt{17}]}^{1/3}$

Let $y=8x^2-x^5$
Hence the required area will be
$={\int }_1^{c}y.dx$
$=[\frac{8x^3}{3}-\frac{x^6}{6}{]}_1^c$
$=(\frac{8c^3}{3}-\frac{c^6}{6})-(\frac{8}{3}-\frac{1}{6})$
$=\frac{16}{3}$
$\therefore \frac{16c^3-c^6}{6}-\frac{15}{6}=\frac{16}{3}$
$\frac{16c^3-c^6}{6}=\frac{32+15}{6}$
$16c^3-c^6=47$
$c^6-16c^3+47=0$
$t^2-16t+47=0$ where $t=c^3$
Hence, $t=\frac{16\pm \sqrt{256-4\left(47\right)}}{2}$
$=8\pm \sqrt{64-47}$
$=8\pm \sqrt{17}$.
Hence, $c=(8\pm \sqrt{17}{)}^{\frac{1}{3}}$.