# how you show integral sign(0-pie/2)f(x)dx < or equal 1? given x*f(x)=sinx

*print*Print*list*Cite

### 1 Answer

You need to prove the following inequality such that:

`int_0^(pi/2) f(x) dx <= cos 1`

The problem provides the information that `x*f(x) = sin x => f(x) = sin x/x.`

Since `sin x/x < sin x` , if `x in (0,pi/2), ` hence `int_0^(pi/2) f(x) dx = int_0^(pi/2)sin x/x dx < int_0^(pi/2) sin x dx`

Evaluating `int_0^(pi/2) sin x dx` yields:

`int_0^(pi/2) sin x dx = -cos x|_0^(pi/2)`

You need to use fundamental theorem of calculus, such that:

`int_0^(pi/2) sin x dx = - cos (pi/2) - (- cos 0)`

`int_0^(pi/2) sin x dx = -0 - (-1) = 1`

**Hence, evaluating `int_0^(pi/2) sin x dx = 1` , yields that the given inequality `int_0^(pi/2) f(x) dx = int_0^(pi/2) sin x/x dx < int_0^(pi/2) sin x dx = 1` holds.**