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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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The correct answer is:
$\frac{129}{10}$
Given, $f(x)=x^{2 / 3}, x \geq 0$ and line $y=x$
$\therefore$ Required area $A=\int_{x=1}^{8}\left(x-x^{2 / 3}\right) d x$
$=\left[\frac{x^{2}}{2}-\frac{3}{5} x^{5 / 3}\right]_{1}^{9}=\left(32-\frac{3}{5} \times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)$
$=32 \times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}$
$=\frac{128+1}{10}=\frac{129}{10}$
$\therefore$ Required area $A=\int_{x=1}^{8}\left(x-x^{2 / 3}\right) d x$
$=\left[\frac{x^{2}}{2}-\frac{3}{5} x^{5 / 3}\right]_{1}^{9}=\left(32-\frac{3}{5} \times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)$
$=32 \times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}$
$=\frac{128+1}{10}=\frac{129}{10}$
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