Solving $\int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx$ I am trying to solve the following integral:
$$\int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx$$
My intuition is that this integral is equal to zero but I am unsure as to which direction to take to prove this. I was thinking binomal expansion but I believe there must be a better way, possibly using summation notation instead.
 A: \begin{align}
\text{Let } u & = 1-x \\
\text{and consequently } x & = 1-u \\
du & = -dx
\end{align}
As $x$ goes from $0$ to $1,$ $u$ goes from $1$ to $0.$
This substitution shows that this integral is $-1$ times this integral. So it is $0.$
 Appendix by the original poster: 
Working with the second integral, his substitution shows that:
$$
\int_0^1x^{300}(1-x)^{700}dx = \int_1^0(1-u)^{300}u^{700}(-du) = \int_0^1u^{700}(1-u)^{300}du
$$
Thus
\begin{align}
& \int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx \\[8pt]
= {} & \int_0^1 x^{700}(1-x)^{300} \, dx - \int_0^1(1-u)^{300}u^{700}\, du = 0
\end{align}
A: Let $1/2 - x \to x^\prime$ to see that the integral is $0$.
Convert the integrand to
$$(x^\prime /2)^{700} (x^\prime /2)^{300} - (x^\prime /2)^{300}(x^\prime /2)^{700}$$
and put in the right limits...
A: Well, solving a more general case we have:
$$\mathcal{I}_\beta\left(\text{n},\text{k}\right):=\int_0^\beta\left(x^\text{n}\left(\beta-x\right)^\text{k}-x^\text{k}\left(\beta-x\right)^\text{n}\right)\space\text{d}x\tag1$$
Let $\text{u}=\beta-x$, so we get $-\text{du}=\text{d}x$, so:
$$\mathcal{I}_\beta\left(\text{n},\text{k}\right)=\int_\beta^0-\left(\left(\beta-\text{u}\right)^\text{n}\text{u}^\text{k}-\left(\beta-\text{u}\right)^\text{k}\text{u}^\text{n}\right)\space\text{du}=$$
$$\int_0^\beta\left(\text{u}^\text{k}\left(\beta-\text{u}\right)^\text{n}-\text{u}^\text{n}\left(\beta-\text{u}\right)^\text{k}\right)\space\text{du}=$$
$$\int_0^\beta\left(x^\text{k}\left(\beta-x\right)^\text{n}-x^\text{n}\left(\beta-x\right)^\text{k}\right)\space\text{d}x\tag2$$
So, we get:
$$\mathcal{I}_\beta\left(\text{n},\text{k}\right)+\mathcal{I}_\beta\left(\text{n},\text{k}\right)=0\tag3$$
A: First let
$$f(x)=x^{700}(1-x)^{300}-x^{300}(1-x)^{700}$$
and see that
$$\begin{aligned}
f\left(\frac{1}{2}+x\right) &= \left(\frac{1}{2}+x\right)^{700}\left(\frac{1}{2}-x\right)^{300}-\left(\frac{1}{2}+x\right)^{300}\left(\frac{1}{2}-x\right)^{700} \\
f\left(\frac{1}{2}-x\right) &= \left(\frac{1}{2}-x\right)^{700}\left(\frac{1}{2}+x\right)^{300}-\left(\frac{1}{2}-x\right)^{300}\left(\frac{1}{2}+x\right)^{700}
\end{aligned}$$
Sum these to get $$f\left(\frac{1}{2}+x\right)+f\left(\frac{1}{2}-x\right)=0$$
$$f\left(\frac{1}{2}+x\right)=-f\left(\frac{1}{2}-x\right)$$
So $f$ is odd with respect to the point $x_0=1/2$ which means
$$\begin{aligned}
\int_0^1 f(x)\,dx&=\int_0^{1/2}f(x)\,dx+\int_{1/2}^1f(x)\,dx \\
&=-\mathcal{J}'+\mathcal{J'}\\
&=0.
\end{aligned}$$
A: We know that the Beta Function is defined by $$\beta(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,\mathrm dx$$
Applying the King rule of integration $$\int_a^b f(x)\,\mathrm dx=\int_a^b f(a+b-x)\,\mathrm dx$$ we can write $$\beta(m,n)=\int_0^1(1-x)^{m-1}x^{n-1}\,\mathrm dx=\beta(n,m)$$
Our integral is of the form
$$\begin{align}I&=\int_0^1x^{m-1}(1-x)^{n-1}\,\mathrm dx-\int_0^1x^{n-1}(1-x)^{m-1}\,\mathrm dx\\&=\beta(m,n)-\beta(n,m)\\&=\beta(m,n)-\beta(m,n)\\&=0\end{align}$$
