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Question Number 184869 by yaslm last updated on 13/Jan/23

Answered by mr W last updated on 20/Jan/23

Commented by mr W last updated on 20/Jan/23

$$y={\left(\frac{x}{b}\right)}^{n}h$$$$A={\int }_0^b{\left(\frac{x}{b}\right)}^{n}hdx=\frac{b^{n+1}h}{(n+1)b^n}=\frac{bh}{n+1}✓$$$$(b-X)A={\int }_0^{b}x{\left(\frac{x}{b}\right)}^{n}hdx=\frac{b^{n+2}h}{(n+2)b^n}=\frac{b^{2}h}{n+2}$$$$(b-X)\frac{bh}{b+1}=\frac{b^{2}h}{n+2}$$$$b-X=\frac{(n+1)b}{n+2}$$$$\Rightarrow X=\frac{b}{n+2}✓$$

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