Question

概率论问题如图所示，第二小问和第三小问的具体的积分步骤。不知道他是怎么积分出来的？

Answer 1

正态分布

f(x) = [1/√(2πσ)] e^.[ -(x-μ)^2/(2σ^2)]

∫(-无穷->+无穷) f(x) dx =1

(2)

E(X)

=k^2.∫(-无穷->+无穷) x^2. e^(-k^2.x^2) dx

=-(1/2) ∫(-无穷->+无穷) x de^(-k^2.x^2)

=-(1/2) [x.e^(-k^2.x^2)]|(-无穷->+无穷) +(1/2)∫(-无穷->+无穷) e^(-k^2.x^2)  dx

=0 +(1/2) ∫(-无穷->+无穷) e^(-k^2.x^2)  dx

μ=0 , σ^2=1/(2k^2)

=(1/2)√(2π) .( k/√2 )  { [1/√(2π).( 1/((√2.k ))] ∫(-无穷->+无穷) e^(-k^2.x^2)  dx }

=(1/2)√(2π) .[1/(√2.k )] (1)

=(1/2) [√π/k]