Question

If $f\left(x\right)=x^{2/3},x\ge 0$. Then,the area of the region enclosed by the curve $y=f\left(x\right)$ and the three lines $y=x,x=1$ and $x=8$ is

• A. $\frac{63}{2}$
• B. $\frac{93}{5}$
• C. $\frac{105}{7}$
• D. $\frac{129}{10}$

Answer 1

The correct option is D $\frac{129}{10}$

Given $f\left(x\right)=x^{2/3},x\ge 0$ and line $y=x$
$\therefore$ Required area $A={\int }_{x=1}^8(x-x^{2/3})dx$
$={[\frac{x^2}{2}-\frac{3}{5}{x}^{5/3}]}_1^8$

$=(32-\frac{3}{5}\times 32)-(\frac{1}{2}-\frac{3}{5})=32\times \frac{2}{5}-\frac{(5-6)}{10}$

$=\frac{64}{5}+\frac{1}{10}=\frac{128+1}{10}=\frac{129}{10}$