The value of $ \int _{0}^{1}x^{99}(1-x)^{100}dx $ is

Question

The value of $\int _{0}^{1}x^{99}(1-x)^{100}dx $ is

Not able to do. I'm trying substituton. But clear failure. Please help.

Answer 1

You can just repeatedly integrate by parts: \begin{align} \int_0^1 x^{99}(1 - x)^{100}\,dx &= {99 \over 101} \int_0^1 x^{98}(1 - x)^{101}\,dx \\ &= {99 \over 101} {98 \over 102} \int_0^1 x^{97}(1 - x)^{102}\,dx \\ &= {99 \over 101} {98 \over 102} {97 \over 103} \int_0^1 x^{96}(1 - x)^{103}\,dx \\ &\cdots \\ &= 99! {100! \over 199!} \int (1 - x)^{199}\,dx \\ &= {99! \,100! \over 200!} \end{align}

Answer 2

$\int_0^1x^{99}(1-x)^{100}dx=B(100,101)=\frac{99!100!}{200!}$

Answer 3

Denote $B(r,s) = \int_0^1 x^{r-1}(1-x)^{s-1}\ dx$. Then your integral is $B(100,101)$, which is just the Beta Function.

So your integral is just $\frac{99! 100!}{200!}$.

Answer 4

Let

$$I(n,m)=\int_0^1 x^n(1-x)^mdx$$ then by integration by parts we get

$$I(n,m)=\frac m{n+1}I(n+1,m-1)$$ and by induction we have

$$I(n,m)=\frac{m!n!}{(m+n)!}I(n+m,0)=\frac{m!n!}{(m+n+1)!}$$

Answer 5

HINT:

Use $I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

$I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx$

Then set $x=\sin^2\theta$

Answer 6

This answer is due to T. Bayes known as Bayes Billiard Balls where he tells nice stories to establish:

$$\int_0^1 {n\choose{k}}x^k(1-x)^{n-k}dx=\frac{1}{n+1}$$

Therefore $$\int_0^1 {199\choose{99}}x^{99}(1-x)^{199-99}dx=\frac{1}{199+1}$$thus $$\int_0^1 x^{99}(1-x)^{100}dx=\frac{99!100!}{199!}\frac{1}{200}=\frac{99!100!}{200!}$$