# Thread: integration by part theorem!

1. ## integration by part theorem!

i need some clarification about the problem below;

$\int_{-1.2}^{3.9} xd[x]$

the only way i understand to solve the above problem is to apply integration by part

that is

$\int_{-1.2}^{3.9} xd[x] + \int_{-1.2}^{3.9} |x|dx = f(b)g(a) - f(a)g(a)$

do i just substitute a as -1.2 and b as 3.9 and that's all?

pls help here

and again how do one disband modulus sign from integration function? asking for general purpose
thanks

2. ## Re: integration by part theorem!

What do you mean by d[x]? Is that the "greatest integer function" or "floor function", $\lfloor x\rfloor$? If so that is a "Stieljes" integral, not a "Riemann" integral. $\lfloor x \rfloor$ changes value only at integer values so the Stieljes integral is just the sum of the function being integrated at -1, 0, 1, 2, and 3. Since here the function is just "x", that would be -1+ 0+ 1+ 2+ 3= 5.

I have no idea why you think you should just add a new integral to it! That certainly has nothing to do with "integration by parts".

3. ## Re: integration by part theorem!

Originally Posted by lawochekel
i need some clarification about the problem below;

$\int_{-1.2}^{3.9} xd[x]$

the only way i understand to solve the above problem is to apply integration by part
I think that you have misunderstood the parts rule.
See this source.

Also have a look at this.

4. ## Re: integration by part theorem!

i want to be sure if i am improving on this, pls check the problem below i attempted

$\int_{-0.3}^{4.5} x^2d([x]) = \int_{-0.3}^{0} x^2d([x]) + \int_{0}^{2} x^2d([x]) + \int_{2}^{4} x^2d([x]) + \int_{4}^{4.5} x^2d([x])$

$=[0-(-1)]0^2+[2-0]2^2+[4-2]4^2+[4-4](4.5)^2 = 0+8+32++0 = 40$

thanks