Question

Prove Corollary$2.1$.

Short answer

Therefore,

$E(ax+b)=aE\left(x\right)+b$

Hence Proved.

Step-by-step solution

Given Information.

$E(ax+b)=aE\left(x\right)+b$

Explanation.

$E(ax+b)=aE\left(x\right)+b$

$E(ax+b)={\int }_0^{\infty }(ax+b)f\left(x\right)dx$

Explanation.

$={\int }_0^{\infty }ax.f\left(x\right)dx+{\int }_0^{\infty }b.f\left(x\right)dx$

$=a{\int }_0^{\infty }x.f\left(x\right)dx+b.{\int }_0^{\infty }f\left(x\right)dx$

$=aE\left(x\right)+b$