Math Help

tarnok

Diabloii.Net Member
Math Help

The following appears in a text book I desperately need to understand by Tuesday:

Integral from z to 1000 of (x-z) * 0.001 dx = 0.0005(1000-z)^2

wtf?

Is there an error there? Am I missing something obvious? Have I completely forgotten integral calculus?

Shouldn't Integral from z to 1000 of (x-z) * 0.001 dx = 0.0005(1000^2 -500z + z^2)?

Help!
 

tarnok

Diabloii.Net Member
Re: Math Help

I'm not actually still in school. I'm trying to figure this out on my own for the actuary exam so I can stop being a teacher.

And it seems I can't remember how to factor. I divided the middle term by 2 instead of 1/2. Correct application of basic arithmetic yields the desired result.

:\
 

SnickerSnack

Diabloii.Net Member
Re: Math Help

Integral from z to 1000 of (x-z) * 0.001 dx = 0.0005(1000-z)^2
Shouldn't Integral from z to 1000 of (x-z) * 0.001 dx = 0.0005(1000^2 -500z + z^2)?
It will be a bit easier to show here if we do a substitution: u=x-z, so du=dx, then

Using the notation, int{integrand, lower limit, upper limit},

int{0.001*(x-z),z,1000}=0.001*int{(x-z),z,1000}
......................................=0.001*int{u,z-z,1000-z}
......................................=0.001*{0.5*u^2,0,1000-z}
......................................=0.001*0.5*(1000-z)^2-0
......................................=0.0005*(1000-z)^2

Though it seems that you figured it out.


 

Spinns

Diabloii.Net Member
Re: Math Help

Well i was looking at the join date of the op which as i see it isn't from 10 years ago
 

krischan

Europe Trade Moderator
Re: Math Help

Yippee, a math help thread. Nerd alarm :D

Shouldn't Integral from z to 1000 of (x-z) * 0.001 dx = 0.0005(1000^2 -500z + z^2)?
As drawing 0.5 out of the bracket (after solving the integral) means multiplying the linear part by 2 instead of dividing, it's 0.0005(1000²-2000z+z²) and (a-b)²=a²-2ab+b²...

Haven't you been here for nearly 10 years? How can you still be in school!?
Maybe he was 8 when he joined :whistling:



 
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