Evaluating $\int\limits_a^b x (b-x)^{n-1} (x-a)^{k-n} dx$

Question

I have an integral:

$$\int\limits_a^b x (b-x)^{n-1} (x-a)^{k-n} \, dx$$

Where $0 \leq a \leq b$ are fixed/known reals and $1 \leq n \leq k$ are fixed/known integers.

At first I thought binomial theorem but I think that requires the variables to have absolute value under $1$.

How can I tackle this?

Answer 1

Start with the following substitution: $$\frac{b-x}{b-a}=t \Rightarrow x=b-(b-a)t\Rightarrow dx=-(b-a)dt$$ This will give us: $$I=\int_{a}^{b} \color{red}{x} \color{blue}{(b-x)^{n-1}} \color{green}{(x-a)^{k-n}} dx$$ $$=(b-a)\int_0^1\color{red}{(b-t(b-a))} \color{blue}{((b-a)t)^{n-1}}\color{green}{((1-t)(b-a))^{k-n}}dt$$ $$=\color{orange}{(b-a)^{k}}\int_0^1 \color{red}{(b-t(b-a))}\color{blue}{t^{n-1}}\color{green}{(1-t)^{k-n}}dt$$ Now split in two parts and use the Beta function to get: $$I=(b-a)^k \left(b\int_0^1 t^{n-1}(1-t)^{k-n}dt-(b-a)\int_0^1 t^n(1-t)^{k-n}dt\right)$$ $$=(b-a)^k \left(bB(n,k-n+1)-(b-a)B(n+1,k-n+1)\right)$$ $$=(b-a)^{k}\left(b\frac{\Gamma(n)\Gamma(k-n+1)}{\Gamma(k+1)}-(b-a)\frac{\Gamma(n+1)\Gamma(k-n+1)}{\Gamma(k+2)}\right)$$ $$=(b-a)^k\left(b\frac{(n-1)!(k-n)!}{k!}-(b-a)\frac{n!(k-n)!}{(k+1)!}\right)$$ $$=(b-a)^k\frac{n!(k-n)!}{k!}\left(\frac{b}{n}-\frac{b-a}{k+1}\right)=\boxed{\frac{(b-a)^k}{\binom{k}{n}}\left(\frac{b}{n}-\frac{b-a}{k+1}\right)}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Set $\ds{x \equiv a + \pars{b - a}t}$: \begin{align} &\bbox[10px,#ffd]{\int_{a}^{b}x\pars{b - x}^{n - 1} \pars{x - a}^{k - n}\,\dd x} \\[5mm] = &\ \int_{0}^{1}\bracks{a + \pars{b - a}t} \bracks{\pars{b - a}\pars{1 - t}}^{n - 1} \bracks{\pars{b - a}t}^{n - k}\pars{b - a}\dd t \\[5mm] = &\ \pars{b - a}^{2n -k}\bracks{% a\int_{0}^{1}t^{n - k}\pars{1 - t}^{n - 1}\dd t + \pars{b - a}\int_{0}^{1}t^{n - k + 1}\pars{1 - t}^{n - 1}\,\dd t} \\[5mm] = &\ \pars{b - a}^{2n -k}\bracks{% a\,{\Gamma\pars{n - k + 1}\Gamma\pars{n} \over \Gamma\pars{2n - k + 1}} + \pars{b - a}{\Gamma\pars{n - k + 2}\Gamma\pars{n} \over \Gamma\pars{2n - k + 2}}} \\[5mm] = &\ \pars{b - a}^{2n -k}\bracks{% a\,{\Gamma\pars{n - k + 1}\Gamma\pars{n} \over \Gamma\pars{2n - k + 1}} + \pars{b - a}{\pars{n - k + 1}\Gamma\pars{n - k + 1}\Gamma\pars{n} \over \Gamma\pars{2n - k + 2}}} \\[5mm] = &\ \bbx{\pars{b - a}^{2n - k}\,\bracks{\pars{a + b}n - bk + b}\, {\Gamma\pars{n - k + 1}\Gamma\pars{n} \over \Gamma\pars{2n - k + 2}}} \end{align}