# Expectation of normal distribution

I know this problem is very easy, finding E(x) from the distribution function. I tried to do it by integration by parts, one way I took x to be my first function and the whole exp term to be my 2nd function, but it didn't work, then I split the exponential term to 2 terms combined one with x as my first function and took the second as the 2nd function but it didn't work either. Can someone help me?

Please provide a detailed proof/answer. Justify all your claims.

See attached file for full problem description.

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#### Solution Preview

Please see the attached file.

.f(x) = __1__ exp-( x - μ )^2

√2πσ^2 2σ^2

α

E(x) = ∫ x __1__ exp-(x - μ)^2 dx

-α √2πσ^2 2σ^2

let ( x - μ ) = t ; dx = dt ; dx = √2σ dt

√2σ √2σ

x = √2σ t + μ

Substituting in the above integral we get

α

∫ (√2σ t + μ) { __1__ exp(- t^2 ) }√2σdt

-α √2πσ^2

α

∫ (√2σ t + μ) { __1__ ...

#### Solution Summary

The expert examines expectations of normal distributions.