Problem source

\begin{questionparts}
\item Show that 
\[
\int_{0}^{1}\left(1+(\alpha-1)x\right)^{n}\,\mathrm{d}x=\frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)}
\]
when $\alpha\neq1$ and $n$ is a positive integer. 
\item Show that if $0\leqslant k\leqslant n$ then the coefficient
of $\alpha^{k}$ in the polynomial 
\[
\int_{0}^{1}\left(\alpha x+(1-x)\right)^{n}\,\mathrm{d}x
\]
is 
\[
\binom{n}{k}\int_{0}^{1}x^{k}(1-x)^{n-k}\,\mathrm{d}x\,.
\]
\item Hence, or otherwise, show that 
\[
\int_{0}^{1}x^{k}(1-x)^{n-k}\,\mathrm{d}x=\frac{k!(n-k)!}{(n+1)!}\,.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
u = 1+(\alpha-1)x: && \int_0^1 (1 + (\alpha - 1)x)^n \d x &= \int_{u=1}^{u=\alpha} u^n \frac{1}{\alpha - 1} \d u \\
&&&= \left [\frac{u^{n+1}}{(n+1)(\alpha-1)} \right]_1^\alpha \\
&&&= \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)}
\end{align*}

\item \begin{align*}
&& \int_0^1 (\alpha x + (1-x))^n \d x &= \int_0^1 \sum_{k=0}^n \binom{n}{k} \alpha^k x^k (1-x)^{n-k} \d x \\
&&&= \sum_{k=0}^n \alpha^k \int_0^1 \binom{n}{k}  x^k (1-x)^{n-k} \d x
\end{align*}

Therefore the coefficient of $\alpha^k$ is $\displaystyle  \int_0^1 \binom{n}{k}  x^k (1-x)^{n-k} \d x$

\item The coefficient of $\alpha^{k}$ in $\displaystyle \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)}$ is $\displaystyle \frac1{n+1}$. Therefore 

\begin{align*}
&& \frac1{n+1} &= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} \d x \\
\Rightarrow && \int_0^1 x^k (1-x)^{n-k} \d x &= \frac{k!(n-k)!}{(n+1)n!} \\
&&&= \frac{k!(n-k)!}{(n+1)!}
\end{align*}
\end{questionparts}