Question

The value of $c$ for which the area bounded by the curve $y=8x^2-x^5$, the lines $x=1,x=c$ and $x-axis$ is $\frac{16}{3}$ sq. units is

• A. $1$
• B. $0$
• C. $-2$
• D. $-1$

Answer 1

The correct option is D $-1$

Given the curve $y=8x^2-x^5$, $x=1$ and $x=c$

now $A={\int }_c^1[8x^2-x^5]$

but given that $A=\frac{16}{3}$

$\frac{16}{3}={\int }_c^1[8x^2-x^5]$

On integration and apply upeer and lower limit we will get

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

On simplification, we will get

$c=-1$