# Integral with variable limit

## Homework Statement

Calculate the integral
0x[sinx] where x belongs to (2nπ,2nπ+π) and [] represents the greatest integer function.

[/b] The attempt at a solution[/b]
Basically i am stuck with interpreting the variable limit otherwise it is an easy question. Any insight on how to interpret the limits is appreciated.

tiny-tim
Homework Helper
hi aim1732! do you mean ∫0x [sinx] dx ?

that must be wrong …

you can't have x (or any function of x) as a limit of integration if x is the variable of integration (ie if it ends "dx") (and anyway, where does the interval (2nπ,2nπ+π) conme into it? )

HallsofIvy
Homework Helper
While it is not very good notation you will sometimes see
$$\int_a^x f(x) dx$$
It is essentially the "anti-derivative" with a specific constant of integration given by the lower limit.

Better notation is to change the 'dummy' index:
$$\int_a^x f(t)dt$$

Here that would be
$$\int_0^x [sin(t)] dt$$

And now all you need to know is that $[sin(t)]= 0$ for $0\le t\le\pi$, $[sin(t)]= -1$ for $\pi< t\le 2\pi$ and then alternates values of 0 (for $2n\pi< t\le (2n+1)\pi$ and -1 (for $(2n+1)\pi\le t< 2(n+1)\pi$).

Oh dear me I forgot the element dx. Of course i mean ∫0x [sinx] dx.

It's a confusing expression but I can't help it if books use it.
So in the give interval
(2nπ,2nπ+π)
is the solution zero?

tiny-tim
Homework Helper
Oh dear me I forgot the element dx. Of course i mean ∫0x [sinx] dx.
It's a confusing expression but I can't help it if books use it.
So in the give interval
(2nπ,2nπ+π)
is the solution zero?
… where x belongs to (2nπ,2nπ+π) and [] represents the greatest integer function.

Basically i am stuck with interpreting the variable limit otherwise it is an easy question. Any insight on how to interpret the limits is appreciated.

Do you mean eg [0.5] =1, [-0.5] = 0 ?

Then in (2nπ,(2n+1)π), [sint] = 1, doesn't it. so that the integral is ∫ 0x dt ?

Do you mean eg [0.5] =1, [-0.5] = 0 ?
I believe it is the ceiling function you are talking about---- slightly different from what i am talking about(the floor function).
Since in the given range of x, [sint] is 0 hence the integration amounts to calculating the area under the curve before the given interval which gives the constant function -nπ--basically the upper limit of integration becomes constant-- but if the interval of x is [(2n+1)π,(2n+2)π] will the integral become a function of x?

tiny-tim
yes, then it'll be constant at -nπ between (2nπ, (2n+1)π), decreasing steadily to -(n+1)π over the (2n+1)π,(2n+2)π) 