Indefinite Integral

[jacks]

Calculation of integral \(\displaystyle \int_{0}^{1} x^{30}.(1-x)^{70}dx\)

[achilleas]

Integration by parts with \(u=(1-x)^{n-k}\) and \(dv=x^k dx\) gives us

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

so

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

since \({n \choose k+1}=\frac{n-k}{k+1}{n \choose k}\),

that is, \(J_{n,k}=J_{n,k+1}\), where \(J_{n,k}={n \choose k}\int_0^1 x^k (1-x)^{n-k} dx= \int_0^1 {n \choose k} x^k (1-x)^{n-k} dx\).

Thus

\(J_{n,k}=J_{n,n}=\int_0^1 x^n dx= \frac{1}{n+1}\)

and so,

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

Hence

\(\int_0^1 x^{30}(1-x)^{70}=\dfrac{1}{101{100 \choose 30}}=\dfrac{70!\cdot 30!}{101!}\)

[jacks]

Thanks achilleas

Can we solve it Using Trig. Substitution.