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

g i v e n t h a t \frac{1}{σ \sqrt{2 π}} \underset{- \infty}{\int^{+ \infty}} e^{- \frac{{(t - μ)}^{2}}{2 σ^{2}}} d t = 1

a n d g (n, u) = \frac{1}{σ \sqrt{2 π}} \underset{- \infty}{\int^{+ \infty}} {(x - u)}^{n} e^{- \frac{{(t - μ)}^{2}}{2 σ^{2}}} d t

a n d f (n, u) = \frac{1}{σ \sqrt{2 π}} \underset{- \infty}{\int^{+ \infty}} {(t - u)}^{n} e^{- \frac{{(t - μ)}^{2}}{2 σ^{2}}} d t

i i . e v a l u a t e f (1, 0)

i i i . e v a l u a t e f (2, 0)

i v . e v a l u a t e f (1, μ)

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

f (n, u) = {(x - u)}^{n} \frac{1}{σ \sqrt{2 π}} \underset{- \infty}{\int^{+ \infty}} e^{- \frac{{(t - μ)}^{2}}{2 σ^{2}}} d t = {(x - u)}^{n}

x, n and u are constant for the 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 .