If f(x)=x^{2/3}, x geq 0. Then, the area of t | vedclass

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(D) Given, $f(x)=x^{2/3}$ and the line $y=x$.For $x \in [1, 8]$, $x \geq x^{2/3}$ because $x^3 \geq x^2$ for $x \geq 1$.The required area $A$ is given

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$\frac{129}{10}$

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(D) Given, $f(x)=x^{2/3}$ and the line $y=x$.For $x \in [1, 8]$, $x \geq x^{2/3}$ because $x^3 \geq x^2$ for $x \geq 1$.The required area $A$ is given by the integral:$A = \int_{1}^{8} (x - x^{2/3}) dx$$= \left[ \frac{x^2}{2} - \frac{3}{5} x^{5/3} ight]_{1}^{8}$$= \left( \frac{8^2}{2} - \frac{3}{5} (8)^{5/3} ight) - \left( \frac{1^2}{2} - \frac{3}{5} (1)^{5/3} ight)$$= \left( \frac{64}{2} - \frac{3}{5} 	imes 32 ight) - \left( \frac{1}{2} - \frac{3}{5} ight)$$= \left( 32 - \frac{96}{5} ight) - \left( \frac{5-6}{10} ight)$$= \left( \frac{160-96}{5} ight) - \left( -\frac{1}{10} ight)$$= \frac{64}{5} + \frac{1}{10} = \frac{128+1}{10} = \frac{129}{10}$.

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

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