Finding the mean and variance of an exponential probability distribution I'm taking a probability theory course, and I'm struggling a bit with gamma and exponential distributions. 
Here's a question that I'm stuck on:
The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula C=100+40Y+3Y^2 relates the cost C of completing this operation to the square of the time to completion. Find the mean and variance of C. 
So I know that for this exponential distribution, beta=E(Y)=10. 
So E(C)=100+40(10)+3E(Y^2)
I'm completely lost on how to find Y^2. Likewise, I don't know how to find the variance V(C) either (I know this involves finding E(Y^3) and E(Y^4)). 
For reference, E(C)=1100 and V(C)=2920000. I just can't get to those numbers.
Thank you very much in advance. 
 A: Let $X$ be standard exponentially distributed, i.e. with parameter $\lambda=1$.
Then it is a nonnegative random variable with PDF $e^{-x}$ for $x>0$.
This allows you to find $\mathbb{E}X^{n}=\int_{0}^{\infty}x^{n}e^{-x}dx=\Gamma\left(n+1\right)=n!$
Here $Y$ is exponentially distributed and can be written as $Y=10X$.
Then $\mathbb{E}Y^{n}=\mathbb{E}\left(10X\right)^{n}=10^{n}\mathbb{E}X^{n}=10^{n}n!$
A: The expectation of $Y^k$ is
$$\int_0^\infty \frac{y^k}{10} e^{-y/10}\,dy.$$
We need $E(Y^k)$ for $k=2$ and (for the variance) for $k=3$ and $k=4$.
We show in detail how to deal with the case $k=2$, using integration by parts.
Let $u=y^2$ and $dv=\frac{1}{10}e^{-y/10}\,dy$. Then $du=2y\,dy$ and we can take $v=-e^{-y/10}$. So our integral is
$$\left.-y^2e^{-y/10}\right|_0^{\infty}+\int_0^\infty 2ye^{-y/10}\,dy.$$
The first term is $0$. For the remaining integral, we could integrate by parts again. But it is easier to note that we know that $\int_0^\infty \frac{y}{10}e^{-y/10}\,dy=10$ (the mean), so the remaining integral is $2(10^2)$. 
The other expectations are handled similarly. For $E(Y^3)$ after one integration by parts step you will be able to reuse the fact just established that $E(Y^2)=2(10^2)$. 
Remark: An easier way is to look up the $k$-th moment of $Y$ about the mean. This information is available in the Wikipedia article on the exponential distribution, and elsewhere. 
