Integral Calculus
86443
\(\int_{0}^{1} x(1-x)^{n} d x=\)
1 \(\frac{4}{(n+1)(n+2)}\)
2 \(\frac{n+3}{(n+1)(n+2)}\)
3 \(\frac{2 \mathrm{n}+3}{(\mathrm{n}+1)(\mathrm{n}+2)}\)
4 \(\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\)
Explanation:
(D) : Given,
\(I=\int_{0}^{1} x(1-x)^{n} d x\)
Let, \(\quad 1-\mathrm{x}=\mathrm{t} \Rightarrow \mathrm{x}=1-\mathrm{t}\)
When,
\(\mathrm{dx}=-\mathrm{dt}\)
\(x=0, t=1\)
\(x=1, t=0\)
Now,
\(I=\int_{1}^{0}-(1-t) t^{n} d x\)
\(=\int_{0}^{1}(\mathrm{t}-1) \mathrm{t}^{\mathrm{n}} \mathrm{dt}\)
\(=\int_{0}^{1} \mathrm{t}^{\mathrm{n}+1} \mathrm{dt}-\int_{0}^{1} \mathrm{t}^{\mathrm{n}} \mathrm{dt}\)
\(=\left[\frac{\mathrm{t}^{\mathrm{n}+2}}{\mathrm{n}+2}\right]_{1}^{0}-\left[\frac{\mathrm{t}^{\mathrm{n}+1}}{\mathrm{n}+1}\right]_{1}^{0}\)
\(=\frac{1}{\mathrm{n}+2}[0-1]-\frac{1}{\mathrm{n}+1}[0-1]\)
\(=\frac{-1}{\mathrm{n}+2}+\frac{1}{\mathrm{n}+1}\)
\(=\frac{-\mathrm{n}-1+\mathrm{n}+2}{(\mathrm{n}+2)(\mathrm{n}+1)}=\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\)
