Question

By using the properties of definite integrals,evaluate the integrals: $\int^1_0 {x(1-x)^n} \, dx$

Answer 1

Answer:

$\bf\int^1_0 {x(1-x)^n}\:dx=\frac{1}{(n+1)(n+2)}$

Step-by-step explanation:

Let

$I=\int^1_0 {x(1-x)^n}\:dx$

Using the properties of definite integrals

$\boxed{\int\limits^a_0\:f(x)\:dx=\int\limits^a_0\:f(a-x)\:dx}$

$I=\int^1_0 {(1-x)(1-(1-x))^n}\:dx$

$I=\int^1_0 {(1-x)(1-1+x)^n}\:dx$

$I=\int^1_0 {(1-x)x^n}\:dx$

$I=\int^1_0 {[x^n-x^{n+1}]}\:dx$

$I= [\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]^1_0$

$I= [\frac{1^{n+1}}{n+1}-\frac{1^{n+2}}{n+2}]-[\frac{0^{n+1}}{n+1}-\frac{0^{n+2}}{n+2}]$

$I= [\frac{1}{n+1}-\frac{1}{n+2}]-[0-0]$

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

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

$\implies\:\boxed{\int^1_0 {x(1-x)^n}\:dx=\frac{1}{(n+1)(n+2)}}$

Answer 2

Answer:

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

Step-by-step explanation:

In this question,

By using the properties of definite integrals we have to evaluate the integrals

I = $\int^1_0 {x(1-x)^n}\:dx$

Using the properties of definite integrals

[tex]\mathbf{{\int\limits^a_0\:f(x)\:dx=\int\limits^a_0\:f(a-x)\:dx} }[/tex]

[tex]I=\int^1_0 {(1-x)(1-(1-x))^n}\:dx [/tex]

[tex]I=\int^1_0 {(1-x)(1-1+x)^n}\:dx [/tex]

$I=\int^1_0 {(1-x)x^n}\:dx$

$I=\int^1_0 {[x^n-x^{n+1}]}\:dx$

[tex]I= [\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]^1_0 [/tex]

[tex]I= [\frac{1^{n+1}}{n+1}-\frac{1^{n+2}}{n+2}]-[\frac{0^{n+1}}{n+1}-\frac{0^{n+2}}{n+2}] [/tex]

$I= [\frac{1}{n+1}-\frac{1}{n+2}]-[0-0]$

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

[tex]I=\frac{1}{(n+1)(n+2)} [/tex]

$\implies\:\mathbf{\int^1_0 {x(1-x)^n}\:dx=\frac{1}{(n+1)(n+2)}}$