[K Sengupta]

Evaluate:

 3
∫  {sin {x}} dx 
-3

where, [x] is defined as the greatest integer ≤ x, and: {x} = x - [x]

This post has been edited by K Sengupta: 03 November 2009 - 05:08 PM

[Freaky]

In a simple integral evaluation...

Spoiler for the answer is...

0, or if you want the absolute value of the area it is 1.997.

But I have a feeling you want more than just that, not quite sure what you mean by "[x] is defined as the greatest integer ≤ x, and: {x} = x - [x]" though.
The way I'm seeing it is that if [x] is the greatest integer ≤ x, that pretty much makes it x, and {x} is 0.

[psychic_mind]

Spoiler for I think

6-6cos(1) or 2.758

This post has been edited by psychic_mind: 03 November 2009 - 07:58 PM

[jerbil]

Spoiler for My Solution

In the range 0 <= x <= 3, the value of [sin x] is zero, except for the instantaneous value of x = pi/2 which does not contribute to the integral. It follows that, in this range the effective value of {sin x} is just sin x.

In the range -3 <= x < 0 the value of [sin x] is -1, so that the integrand is sin x - (-1), or sin x + 1.

The integral of sin x from -3 < x < 3 is zero. Hence the total integral has the value 3.

Have I made a mistake somewhere? It is all more confusing than I originally thought.

[Tuckleton]

Spoiler for I agree with PM

The Function {x} = x - [x] is basically a kind of reverse truncation where you remove everything before the decimal point. You basically get ∫sin(x)dx from [0 to 1) 6 times.

[jerbil]

Spoiler for Mathematica Result

Mathematica integrates sin x - Floor(sin x) between -3 and 3 as 3.

I have just checked it. See my post #4.

This post has been edited by jerbil: 04 November 2009 - 03:14 PM

[jerbil]

Spoiler for Spotted my Error

I thought that the integrand was {sin(x)}, not {sin{x}}. Apologies all.