Question

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

• A. $-1$
• B. $3$
• C. $2$
• D. $\sqrt[3]{38-\sqrt{17}}$

Answer 1

The correct option is D $\sqrt[3]{38-\sqrt{17}}$

By the given condition

${\int }_1^k(8x^2-x^5)dx=\frac{16}{3}\Rightarrow {[\frac{8x^3}{3}-\frac{x^6}{6}]}_1^k=\frac{16}{3}$

$\Rightarrow \frac{8}{3}(k^3-1)-\frac{1}{6}(k^6-1)=\frac{16}{3}$

$\Rightarrow (\frac{8k^3}{3}-\frac{k^6}{6})-(\frac{8}{3}-\frac{1}{6})=\frac{16}{3}$

$\Rightarrow 16k^3-k^6-15=32\Rightarrow k^6-16k^3+47=0$

$\Rightarrow k^3=8\pm \sqrt{17}\Rightarrow k=\sqrt[3]{38-\sqrt{17}}$