Homework Help

Explain please how is demonstrate definite integral from 1 to pie/2 f(x) < cos 1?...

user profile pic

ruals | Salutatorian

Posted May 24, 2013 at 5:58 PM via web

dislike 2 like

Explain please how is demonstrate definite integral from 1 to pie/2 f(x) < cos 1?

xf(x)= sinx

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 24, 2013 at 6:49 PM (Answer #1)

dislike 1 like

You need to prove that `int_1^(pi/2) f(x)dx < cos 1` , hence, you need to find `f(x)` , using the information provided by the problem, such that:

`x*f(x) = sin x => f(x) = (sin x)/x`

Replacing `(sin x)/x` for `f(x)` yields:

`int_1^(pi/2) (sin x)/x dx < cos 1`

Since `x in [1,pi/2]` , hence, the following inequality holds:

`(sin x)/x < (sin x)/1`

Integrating both sides, yields:

`int_1^(pi/2) (sin x)/x dx < int_1^(pi/2) (sin x)/1 dx`

Evaluating the definite integral yields:

`int_1^(pi/2) (sin x)/1 dx = -cos x|_1^(pi/2)`

Using the fundamental theorem of calculus yields:

`int_1^(pi/2) (sin x)/1 dx = -cos (pi/2) - (- cos 1)`

Since `cos (pi/2) = 0` yields:

`int_1^(pi/2) (sin x)/1 dx = cos 1`

Replacing `cos 1` for `int_1^(pi/2) (sin x)/1 dx`   in inequality considered yields:

`int_1^(pi/2) (sin x)/x dx < int_1^(pi/2) (sin x)/1 dx = cos 1`

Hence, testing if the inequality` int_1^(pi/2) (sin x)/x dx < cos 1` holds, using the inequality `(sin x)/x < (sin x)/1` , `x in [1,pi/2]` yields `int_1^(pi/2) (sin x)/x dx < int_1^(pi/2) (sin x)/1 dx = cos 1` valid.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes