Question

integrate a Abs(f(x)) function

Answer 1

\[f'(x)\frac{f(x)}{|f(x)|}\]

Answer 2

oh, integrate ...

Answer 3

we can always do it in pieces

Answer 4

cos(x) >= 0 from -pi/2 to pi/2

or from: 0 to pi/2, then pi/2 to 3pi/2, etc ..

depending on where L lands we have multiples and a half of the integral from pi/2 to 3pi/2 and then add on the excess from L

Answer 5

\[\int_{0}^{pi/2}+k\int_{pi/2}^{3pi/2}+\int_{(2k+1)pi/2}^{L}\]

Answer 6

No need to include the Pi, right?

Answer 7

pi is a number, its going to be required

Answer 8

there is a pi in the function already. that's what I meant

Answer 9

so, pis a number, its part of the function. i see no reason to just throw it out the window.

let

u = pi x/L

du = pi/L dx

dx = L/pi du

\[\frac{bL}{\pi}\int_{u(0)}^{u(L)}|\cos(u)|~du\]

Answer 10

when x=0, u=0

when x=L, u=pi

Answer 11

\[\frac{bL}{\pi}\int_{u(0)=0}^{u(L)=\pi}|\cos(u)|~du\]

Answer 12

that would be \[\frac{ 2bL }{ \pi }\]

Answer 13

is it
\[\int\limits_{0}^{1/2L}b*|\cos(\frac{ \pi*x }{ L })|+\int\limits_{1/2L}^{L}b*|\cos(\frac{ \pi*x }{ L })|\]

Answer 14

\[\int\limits_{0}^{1/2L}b*|\cos(\frac{ \pi*x }{ L })|dx+\int\limits_{1/2L}^{L}b*|\cos(\frac{ \pi*x }{ L })|dx\]

Answer 15

with the change in variable the setup becomes quite simple ... the stuff posted at the start of this was just me trying to get my mind in order .

|cos(u)| is a graph of humps



now the area under the curve from 0 to pi, is the same as the area under the curve from -pi/2 to pi/2. what im getting is that cos(u) = |cos(u)| between -pi/2 and pi/2

so our integration is equal to (if the absolute value part is messing with you):
\[\int_{-\pi/2}^{\pi/2}\frac{bL}{\pi}cos(u)~du\]
\[\frac{bL}{\pi}\int_{-\pi/2}^{\pi/2}cos(u)~du\]
\[\left.\frac{bL}{\pi}sin(u)\right|_{-\pi/2}^{\pi/2}\]