Homework Help

Show `int_0^(2pi )f^n (x)dx` less 2^n `pi`   f(x )=1+sin x 

user profile pic

ruals | (Level 1) Salutatorian

Posted June 27, 2013 at 4:01 PM via web

dislike 1 like

Show `int_0^(2pi )f^n (x)dx`

less 2^n `pi`

 

f(x )=1+sin x 

1 Answer | Add Yours

Top Answer

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 27, 2013 at 5:24 PM (Answer #1)

dislike 1 like

You need to check the following inequality, such that:

`int_0^(2pi) f^n(x) dx < 2^n*pi`

Replacing ` (1 + sin x)` for `f(x)` yields:

`int_0^(2pi) (1 + sin x)^n dx < 2^n*pi`

You may use the following property of integral such that:

`int_0^(2pi) (1 + sin x)^n dx = int_0^(pi) (1 + sin x)^n dx + int_pi^(2pi) (1 + sin x)^n dx`

You should notice that you may write `int_pi^(2pi) (1 + sin x)^n dx` such that:

`int_pi^(2pi) (1 + sin x)^n dx = int_0^(pi) (1 + sin (x + pi))^n dx`

`sin (x + pi) = sin x*cos pi + sin pi*cos x`

Since `cos pi = -1` and `sin pi = 0` yields:

`sin (x + pi) = - sin x`

`int_pi^(2pi) (1 + sin x)^n dx = int_0^(pi) (1 - sin x)^n dx`

Replacing `int_0^(pi) (1 - sin x)^n dx` for `int_pi^(2pi) (1 + sin x)^n dx` yields:

`int_0^(2pi) (1 + sin x)^n dx = int_0^(pi) (1 + sin x)^n dx + int_0^(pi) (1 - sin x)^n dx`

Using the property of linearity of integral yields:

`int_0^(2pi) (1 + sin x)^n dx = int_0^(pi)((1 + sin x)^n + (1 - sin x)^n) dx`

Using binomial theorem yields:

`(1 + sin x)^n + (1 - sin x)^n = sum_(k=1)^n C_n^k sin^k x + sum_(k=1)^n (-1)^k C_n^k sin^k x`

`(1 + sin x)^n + (1 - sin x)^n = 2sum_(k=1)^(n/2) C_n^(2k) sin^(2k) x`

Since `sin^(2k) x < 1` yields:

`(1 + sin x)^n + (1 - sin x)^n = 2sum_(k=1)^(n/2) C_n^(2k) sin^(2k) x < 2sum_(k=1)^(n/2) C_n^(2k) = 2*2^(n - 1)`

`(1 + sin x)^n + (1 - sin x)^n < 2^(1+n-1)`

`(1 + sin x)^n + (1 - sin x)^n < 2^n`

Integrating both sides yields:

`int_0^(pi)((1 + sin x)^n + (1 - sin x)^n) dx <int_0^(pi) 2^n dx`

`int_0^(pi) 2^n dx = 2^n*x|_0^pi`

Using the fundamental theorem of calculus, yields:

`int_0^(pi) 2^n dx = 2^n*(pi - 0)`

`int_0^(pi) 2^n dx = 2^n*pi`

Replacing `2^n*pi` for `int_0^(pi) 2^n dx` yields:

`int_0^(pi)((1 + sin x)^n + (1 - sin x)^n) dx < 2^n*pi`

Replacing `int_0^(2pi) (1 + sin x)^n dx` for `int_0^(pi)((1 + sin x)^n + (1 - sin x)^n) dx` yields:

`int_0^(2pi) (1 + sin x)^n dx < 2^n*pi`

Hence, testing if the inequality `int_0^(2pi) (1 + sin x)^n dx < 2^n*pi` holds, using the properties of integrals and binomial theorem, yields that the statement is valid.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes