Calculating $\int_{0}^{1} (x - x^2)^n dx$. I've been struggling with this question for some time now,
Ive tried doing it the way that $\int_{0}^{1} (1 - x^2)dx$ is done but was unsucessful.
Any hints are appreciated!
 A: Hints. You can use the binomial theorem.
A: If you want to use a bit of prior knowledge:
$I=\int_0^1 (x-x^2)^n dx=\int_0^1 x^n(1-x)^ndx = B(n+1,n+1)$
where $B(x,y)$ is the beta function https://en.wikipedia.org/wiki/Beta_function .
Now using the relationship between the beta and gamma function:
$B(n+1,n+1)=\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}=\frac{n!n!}{(2n+1)!}= \frac{1}{2n+1}{{2n}\choose {n}}^{-1}$ 
So that:
$$I=\frac{1}{2n+1}{{2n}\choose {n}}^{-1}$$
Using the Binomial theorem instead:
$(x-x^2)^n=\sum_{k=0}^{n} {{n}\choose{k}}(-1)^kx^{2k}x^{n-k}$
And integrating each term of the summation:
$$I=\sum_{k=0}^{n} {{n}\choose{k}}(-1)^k\frac{1}{(n+k+1)}$$
We have than two ways of expressing the same quantity. We have therefore as a bonus a nontrivial (for me) identity:
$$\frac{1}{2n+1}{{2n}\choose {n}}^{-1}=\sum_{k=0}^{n} {{n}\choose{k}}(-1)^k\frac{1}{(n+k+1)}$$
Hope to not have messed the calculations.. Maybe somebody can prove the identity in a more direct way...
A: If $Re(n)>-1$ and $Re(m)>-1$, then
$$
\int^{1}_{0}x^n(1-x)^mdx=\frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(n+m+2)}
$$
A: Some knowledge first:
$$B(x+1,y+1)=\int_0^1t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}$$
now if we look at your integral:
$$I=\int_0^1(x-x^2)^ndx=\int_0^1\left[x(1-x)\right]^ndx=\int_0^1x^n(1-x)^ndx=B(n+1,n+1)$$
$$I=\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}$$
now if n is an integer we get:
$$\Gamma(n+1)^2=(n!)^2$$
$$\Gamma(2n+2)=(2n+1)!$$
so we get:
$$I=\frac{(n!)^2}{(2n+1)!}$$
This will tend to zero very quickly as n increases. This can be seen by representing it as:
$$I=\frac{1}{2n+1}{^{2n}C_n}^{-1}$$
so it is always less than 1
A: So I finally found and answer how to evaluate this integer in "elementary" way.
Obviously: $$\int_{0}^{1} (x - x^2)^n dx = \int_{0}^{1} x^n(1 - x)^n dx$$ Lets integrate that by partial derivatives. 
$$\int_{0}^{1} x^n(1 - x)^n dx = |\frac{x^n+1}{n+1} (1 - x)^n|_{0}^{1} + \frac{n}{n+1}\int_{0}^{1} \frac{x^{n+1}}{n+1}(1 - x)^{n-1} dx$$
First term is $0$, and from the second one we take partial derivatives the same way we did with the first one. After n number of iterations of partial derivatives we get: $${2n \choose n}^{-1}\int_{0}^{1}x^{2n}dx$$
Which is the same as other answers.
