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Question Number 697 by 123456 last updated on 02/Mar/15

given that (1/(σ(√(2π))))∫_(−∞) ^(+∞) e^(−(((t−μ)^2 )/(2σ^2 ))) dt=1  and g(n,u)=(1/(σ(√(2π)))) ∫_(−∞) ^(+∞) (x−u)^n e^(−(((t−μ)^2 )/(2σ^2 ))) dt  and f(n,u)=(1/(σ(√(2π))))∫_(−∞) ^(+∞) (t−u)^n e^(−(((t−μ)^2 )/(2σ^2 ))) dt  ii. evaluate f(1,0)  iii. evaluate f(2,0)  iv. evaluate f(1,μ)

$${given}\:{that}\:\frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt}=\mathrm{1} \\ $$$${and}\:{g}\left({n},{u}\right)=\frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\:\underset{−\infty} {\overset{+\infty} {\int}}\left({x}−{u}\right)^{{n}} {e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt} \\ $$$${and}\:{f}\left({n},{u}\right)=\frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\underset{−\infty} {\overset{+\infty} {\int}}\left({t}−{u}\right)^{{n}} {e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt} \\ $$$${ii}.\:{evaluate}\:{f}\left(\mathrm{1},\mathrm{0}\right) \\ $$$${iii}.\:{evaluate}\:{f}\left(\mathrm{2},\mathrm{0}\right) \\ $$$${iv}.\:{evaluate}\:{f}\left(\mathrm{1},\mu\right) \\ $$

Commented by prakash jain last updated on 01/Mar/15

f(n,u)=(x−u)^n (1/(σ(√(2π))))∫_(−∞) ^(+∞) e^(−(((t−μ)^2 )/(2σ^2 ))) dt=(x−u)^n   x, n and u are constant for the integration.

$${f}\left({n},{u}\right)=\left({x}−{u}\right)^{{n}} \frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt}=\left({x}−{u}\right)^{{n}} \\ $$$${x},\:{n}\:\mathrm{and}\:{u}\:\mathrm{are}\:\mathrm{constant}\:\mathrm{for}\:\mathrm{the}\:\mathrm{integration}. \\ $$

Answered by prakash jain last updated on 01/Mar/15

f(1,0)=x  f(2,0)=x^2   f(1,μ)=(x−μ)  x,n and μ are constants in the integral.

$${f}\left(\mathrm{1},\mathrm{0}\right)={x} \\ $$$${f}\left(\mathrm{2},\mathrm{0}\right)={x}^{\mathrm{2}} \\ $$$${f}\left(\mathrm{1},\mu\right)=\left({x}−\mu\right) \\ $$$${x},{n}\:\mathrm{and}\:\mu\:\mathrm{are}\:\mathrm{constants}\:\mathrm{in}\:\mathrm{the}\:\mathrm{integral}. \\ $$

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