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Anonymous Mon 22 Jul 00:21:47 2019 No.10826127 View Reply Original Report

Quoted By: >>10826154 >>10826167 >>10826174

I have to be honest, I've taken Calc III and Differential Equations, and I have no idea how the author got line 2. What am I missing? I don't know how to work with the fractional part function in a calculus setting

Anonymous

Anonymous Mon 22 Jul 2019 00:34:25 No.10826154 Report

Quoted By: >>10826156

>>10826127
lol brainlet nigger

taking baby calc without having to do proofs means nothing

Anonymous

Anonymous Mon 22 Jul 2019 00:35:28 No.10826156 Report

Quoted By:

>>10826154
Can you at least provide something constructive and not just shitpost?

Anonymous

Anonymous Mon 22 Jul 2019 00:37:11 No.10826161 Report

Quoted By:

Integration by parts bro

Anonymous

Anonymous Mon 22 Jul 2019 00:41:03 No.10826167 Report

Quoted By:

>>10826127
the fraction doesn't effect anything? it just carries through?

Anonymous

Anonymous Mon 22 Jul 2019 00:43:25 No.10826174 Report

Quoted By: >>10826179

>>10826127
On the interval $[k, k+1]$ :
$(x - \text{floor}(x) ) - \frac{1}{2} = Ax+B$
Boundary conditions:
$kA + B = - \frac{1}{2}$
$(k+1) A + B = \frac{1}{2}$

Solving:
$A = 1$
$B = -k - \frac{1}{2}$

On the given interval, the integral becomes:
$\int_{k}^{k+1} [(x - \text{floor}(x)) - \frac{1}{2} ] f'(x) dx = \int_{k}^{k+1} [x- (k+ \frac{1}{2})] f'(x) dx$

The rest in just integration by parts .

Anonymous

Anonymous Mon 22 Jul 2019 00:44:47 No.10826179 Report

Quoted By:

>>10826174
Thank you very much

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