Question

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

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

Answer 1

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

The graph is shown in image

intersection point of $y=x^{\frac{2}{3}}$ and $y=x$ are $(0,0)$ and $(1,1)$

but area enclosed by $f\left(x\right)$ ,$y=x$ ,$x=1$ and $x=8$ is

now $A={\int }_1^8[x]-{\int }_1^8[x^{\frac{2}{3}}]$

$A=[\frac{x^2}{2}]{|}_1^8-[\frac{3x^{\frac{5}{3}}}{5}]{|}_1^8$

$A=\frac{63}{2}-\frac{93}{5}sq.unit$

$A=\frac{129}{10}sq.unit$