Question:

By using the properties of definite integrals, evaluate the integral: 01x(1x)ndx\int^1_0x(1-x)^n\,dx

Updated On: Oct 7, 2023
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Solution and Explanation

Let I=01x(1x)ndx\int_{0}^{1} x(1-x)^n \,dx

∴I=01(1x)(1(1x))ndx\int_{0}^{1} (1-x)(1-(1-x))^n \,dx

=01x(1x)(x)ndx\int_{0}^{1} x(1-x) (x)^n \,dx

=01(xnxn+1)dx=∫^1_0(x^n-x^{n+1})dx

=[xn+1n+1xn+2n+2]     (0aƒ(x)dx=0aƒ(ax)dx)=[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]\,\,\,\,\, (∫^a_0ƒ(x)dx=∫^a_0ƒ(a-x)dx)

=[1n+11n+2][\frac{1}{n+1}-\frac{1}{n+2}]

=(n+2)(n+1)(n+1)(n+2)\frac{(n+2)-(n+1)}{(n+1)(n+2)}

=1(n+1)(n+2)\frac{1}{(n+1)(n+2)}

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