Menu Close

Question-184869




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=((x/b))^n h  A=∫_0 ^b ((x/b))^n hdx=((b^(n+1) h)/((n+1)b^n ))=((bh)/(n+1)) ✓  (b−X)A=∫_0 ^b x((x/b))^n hdx=((b^(n+2) h)/((n+2)b^n ))=((b^2 h)/(n+2))  (b−X)((bh)/(b+1))=((b^2 h)/(n+2))  b−X=(((n+1)b)/(n+2))  ⇒X=(b/(n+2)) ✓
$${y}=\left(\frac{{x}}{{b}}\right)^{{n}} {h} \\ $$$${A}=\int_{\mathrm{0}} ^{{b}} \left(\frac{{x}}{{b}}\right)^{{n}} {hdx}=\frac{{b}^{{n}+\mathrm{1}} {h}}{\left({n}+\mathrm{1}\right){b}^{{n}} }=\frac{{bh}}{{n}+\mathrm{1}}\:\checkmark \\ $$$$\left({b}−{X}\right){A}=\int_{\mathrm{0}} ^{{b}} {x}\left(\frac{{x}}{{b}}\right)^{{n}} {hdx}=\frac{{b}^{{n}+\mathrm{2}} {h}}{\left({n}+\mathrm{2}\right){b}^{{n}} }=\frac{{b}^{\mathrm{2}} {h}}{{n}+\mathrm{2}} \\ $$$$\left({b}−{X}\right)\frac{{bh}}{{b}+\mathrm{1}}=\frac{{b}^{\mathrm{2}} {h}}{{n}+\mathrm{2}} \\ $$$${b}−{X}=\frac{\left({n}+\mathrm{1}\right){b}}{{n}+\mathrm{2}} \\ $$$$\Rightarrow{X}=\frac{{b}}{{n}+\mathrm{2}}\:\checkmark \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *