$X$ has a Gamma distribution with parameters $\lambda$ and $\alpha$. Find $E(X^r)$ $X$ has a Gamma distribution with parameters $\lambda$ and $\alpha$. I must find $E(X^r)$ and $r$ is a positive integer.
How can I do this?
I am guessing I have to use the Gamma function but I don't know how to do this?
 A: You leave it to be inferred by us based only on conventions that $\alpha$, rather than $\lambda$ is the shape parameter.  There's also the question of whether $\lambda$ is supposed to be the intensity parameter, so that the distribution is
$$
\frac{1}{\Gamma(\alpha)}\cdot (\lambda x)^{\alpha-1} e^{-\lambda x}\,(\lambda\,dx)\text{ for }x>0
$$
or its reciprocal, the scale parameter, so that the distribution is
$$
\frac{1}{\Gamma(\alpha)}\cdot (x/\lambda)^{\alpha-1} e^{-x/\lambda}\,(dx/\lambda)\text{ for }x>0.
$$
I'm going to guess that you mean the first of these two alternatives.
Once you understand the integral that defines the Gamma function, you're almost done.  You have
$$
\begin{align}
\mathbb E(X^r) & = \int_0^\infty x^r \frac{1}{\Gamma(\alpha)}\cdot (\lambda x)^{\alpha-1} e^{-\lambda x}\,(\lambda\,dx) \\[12pt]
& = \frac{1}{\Gamma(\alpha)} \cdot\frac{1}{\lambda^r} \int_0^\infty (\lambda x)^{r+\alpha-1} e^{-\lambda x}\,(\lambda\,dx) \\[12pt]
& = \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{\lambda^r} \int_0^\infty u^{r+\alpha-1} e^{-u}\,du \\[12pt]
& = \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{\lambda^r}\cdot\Gamma(r+\alpha). \tag1
\end{align}
$$
This bears simplification.  We have
$$
\begin{align}
\Gamma(r+\alpha) & = (r-1)\Gamma(r-1+\alpha)= (r-1)(r-2)\Gamma(r-2+\alpha)=\cdots \\[12pt]
& \cdots=(r+\alpha-1)(r+\alpha-2)(r+\alpha-3)\cdots(r+\alpha-r)\Gamma(\alpha).
\end{align}
$$
Then you can reduce the fraction in $(1)$.
