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Question Number 64384 by aliesam last updated on 17/Jul/19

Answered by MJS last updated on 17/Jul/19

$$(y=x)\cap (y=\frac{x}{8})=\left(\begin{array}{c}0\\ 0\end{array}\right)$$$$(y=x)\cap (y=\frac{1}{x^2})=\left(\begin{array}{c}1\\ 1\end{array}\right)$$$$[x=\frac{1}{x^2}\Rightarrow x^3=1\Rightarrow x=1]$$$$(y=\frac{x}{8})\cap (y=\frac{1}{x^2})=\left(\begin{array}{c}2\\ \frac{1}{4}\end{array}\right)$$$$[\frac{x}{8}=\frac{1}{x^2}\Rightarrow x^3=8\Rightarrow x=2]$$$$$$$$\underset{0}{\stackrel{1}{\int }}xdx+\underset{1}{\stackrel{2}{\int }}\frac{dx}{x^2}-\underset{0}{\stackrel{2}{\int }}\frac{x}{8}dx={\left[\frac{x^2}{2}\right]}_0^1+{[-\frac{1}{x}]}_1^2-{\left[\frac{x^2}{16}\right]}_0^2=$$$$=(\frac{1}{2}-0)+(-\frac{1}{2}+\frac{1}{1})-(\frac{1}{2}-0)=\frac{1}{2}$$

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