How to calculate MEAN of exponential distribution?

Question

$ f(x) = \begin{cases} \frac15 e^{(-\frac15x)}, x>0 \\ 0, \text{elsewhere}\\ \end{cases}$

How to calculate $E[(X+5)]$ and $E[(X+5)^2]$ ?

Thanks a lot.

Answer 1

Hints: for $E[X+5]$ use the linearity of expectation. What is $E[5]$?

For $E[(X+5)^2]$ you can go back to the definition $$E[(X+5)^2]=\displaystyle \int_0^{\infty} \dfrac{(x+5)^2\exp(\frac{-x}{5})\; dx}{5}$$ or you can expand out $E[(X+5)^2]=E[X^2]+2E[X]E[5]+E[5^2]$

Answer 2

If you know that $$ \int_0^\infty u^2 e^{-u} \, du = 2, $$ then you can write $$ \int_0^\infty x^2 e^{-x/5}\, \frac{dx}{5} = 5^2 \int_0^\infty \left(\frac x 5\right)^2 e^{-x/5} \, \frac{dx}{5} = 5^2\int_0^\infty u^2 e^{-u} \, du = 5^2\cdot 2. $$ That gets you $E(X^2)$. And $E((X+5)^2) = E(X^2) + 10E(X) + 5^2$, etc.