Question Number 19895 by virus last updated on 17/Aug/17
Commented by virus last updated on 17/Aug/17
$${please}\:{solve}\:{this} \\ $$
Answered by mrW1 last updated on 18/Aug/17
![c=((∫_0 ^( a) f(x)xdx)/(∫_0 ^( a) f(x)dx))=((k∫_0 ^( a) x^(n+1) dx)/(k∫_0 ^( a) x^n dx))=((n+1)/(n+2))×(([x^(n+2) ]_0 ^a )/([x^(n+1) ]_0 ^a )) =((n+1)/(n+2))×(a^(n+2) /a^(n+1) )=((n+1)/(n+2))a=a−(a/(n+2)) d=((∫_0 ^( a) f(x)((f(x))/2)dx)/(∫_0 ^( a) f(x)dx))=((k^2 ∫_0 ^( a) x^(2n) dx)/(2k∫_0 ^( a) x^n dx))=(k/2)×((n+1)/(2n+1))×(([x^(2n+1) ]_0 ^a )/([x^(n+1) ]_0 ^a )) =(k/2)×((n+1)/(2n+1))×a^n =((n+1)/(2(2n+1)))b n=0: rectangle, c=a−(a/2)=(a/2), d=(b/2) n=1: triangle, c=a−(a/3)=((2a)/3), d=(b/3) n=2: parabola, c=a−(a/4)=((3a)/4), d=(3/(10))b .....](Q19899.png)
$$\mathrm{c}=\frac{\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{xdx}}{\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}}=\frac{\mathrm{k}\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{x}^{\mathrm{n}+\mathrm{1}} \mathrm{dx}}{\mathrm{k}\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{x}^{\mathrm{n}} \mathrm{dx}}=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{2}}×\frac{\left[\mathrm{x}^{\mathrm{n}+\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{a}} }{\left[\mathrm{x}^{\mathrm{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{a}} } \\ $$$$=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{2}}×\frac{\mathrm{a}^{\mathrm{n}+\mathrm{2}} }{\mathrm{a}^{\mathrm{n}+\mathrm{1}} }=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{2}}\mathrm{a}=\mathrm{a}−\frac{\mathrm{a}}{\mathrm{n}+\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{d}=\frac{\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\mathrm{dx}}{\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}}=\frac{\mathrm{k}^{\mathrm{2}} \int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{x}^{\mathrm{2n}} \mathrm{dx}}{\mathrm{2k}\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{x}^{\mathrm{n}} \mathrm{dx}}=\frac{\mathrm{k}}{\mathrm{2}}×\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}+\mathrm{1}}×\frac{\left[\mathrm{x}^{\mathrm{2n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{a}} }{\left[\mathrm{x}^{\mathrm{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{a}} } \\ $$$$=\frac{\mathrm{k}}{\mathrm{2}}×\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}+\mathrm{1}}×\mathrm{a}^{\mathrm{n}} =\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}\left(\mathrm{2n}+\mathrm{1}\right)}\mathrm{b} \\ $$$$\mathrm{n}=\mathrm{0}:\:\mathrm{rectangle},\:\mathrm{c}=\mathrm{a}−\frac{\mathrm{a}}{\mathrm{2}}=\frac{\mathrm{a}}{\mathrm{2}},\:\mathrm{d}=\frac{\mathrm{b}}{\mathrm{2}} \\ $$$$\mathrm{n}=\mathrm{1}:\:\mathrm{triangle},\:\mathrm{c}=\mathrm{a}−\frac{\mathrm{a}}{\mathrm{3}}=\frac{\mathrm{2a}}{\mathrm{3}},\:\mathrm{d}=\frac{\mathrm{b}}{\mathrm{3}} \\ $$$$\mathrm{n}=\mathrm{2}:\:\mathrm{parabola},\:\mathrm{c}=\mathrm{a}−\frac{\mathrm{a}}{\mathrm{4}}=\frac{\mathrm{3a}}{\mathrm{4}},\:\mathrm{d}=\frac{\mathrm{3}}{\mathrm{10}}\mathrm{b} \\ $$$$..... \\ $$
Commented by virus last updated on 18/Aug/17
$${what}\:{is}\:{with}\:{respect}\:{to}\:{y}\:{axis}???? \\ $$
Commented by virus last updated on 18/Aug/17
$${thank}\:{you}\:{sir} \\ $$
Commented by mrW1 last updated on 18/Aug/17
$$\mathrm{i}\:\mathrm{have}\:\mathrm{added}. \\ $$