# integration and the greatest function help exam in 7 hrs

• Dec 7th 2006, 08:27 AM
myoplex11
integration and the greatest function help exam in 7 hrs
integrate f(x) from 0 to n. where n is a positive integer and f(x) is the greatest integer function.
I Know that i have to use the definiton of greatest integer function and break it into sum of the integrals how do i do this.
• Dec 7th 2006, 08:32 AM
AfterShock
Quote:

Originally Posted by myoplex11
integrate f(x) from 0 to n. where n is a positive integer and f(x) is the greatest integer function.
I Know that i have to use the definiton of greatest integer function and break it into sum of the integrals how do i do this.

F(b) - F(a) is the signed area, where a and b are the limits of integration (from a to b).

In this case, your b is n and 0 is a. Do you have a specific problem?
• Dec 7th 2006, 08:42 AM
myoplex11
this is the problem we had on our practice exam
• Dec 7th 2006, 08:51 AM
ThePerfectHacker
Quote:

Originally Posted by myoplex11
integrate f(x) from 0 to n. where n is a positive integer and f(x) is the greatest integer function.
I Know that i have to use the definiton of greatest integer function and break it into sum of the integrals how do i do this.

Okay.

A defined function on a closed interval is Riemann integrable if and only if it is continous almost everywhere.

We we need to find,
$\displaystyle \int_0^n [ x ] dx$
Since it is continous almost everywhere we can use the subdivision rule,
$\displaystyle \sum_{k=1}^n \int_{k-1}^k [x] dx$
But,
$\displaystyle \int_{k-1}^k [x] dx =k-1$
Because in that $\displaystyle [x]=k-1$ on this interval.

Thus,
$\displaystyle \sum_{k=1}^n k-1=\frac{n(n-1)}{2}$
• Dec 7th 2006, 08:52 AM
ThePerfectHacker
Quote:

Originally Posted by AfterShock
F(b) - F(a) is the signed area, where a and b are the limits of integration (from a to b).

In this case, your b is n and 0 is a. Do you have a specific problem?

That theorem does not work. The fundamental theorem of calculus only applies when a function is countinous on the closed interval.
• Dec 7th 2006, 08:52 AM
Plato
If you draw a graph for say n=8, you will see a series of seven ‘stair steps’.
The area under the ‘stairs’ is the sum of the areas of seven rectangles.
So 1+2+3…+7. To do it for n we get a sum of $\displaystyle \frac{{\left( {n - 1} \right)n}}{2}.$