ntegrating w/ abs value?? (1 Viewer)

[jkwii]

I = S (2- |x|) dx from 3 to -3.. do u use the graph?? like area??

 

[Js^-1]

hmmm...No clue. Its not odd, so its not automatically zero. Maybe you have to integrate the parts seperately and add them? Like for x>0 and x<0? Integrating y=2-x from 0 to 3 and y=x-2 from -3 to 0. I have no idea if thats right or not, but thats how i'd try to do it.

 

[conics2008]

use the fact that |x| = y=x and y = -x above the x axis

 

[Templar]

Break the integral into 2 and |x|, and use the fact |x| is even

 

[conics2008]

Here I will list the steps for you

i) draw 2-|x|
ii) You will notice that its made up of two equations y=2-x and y=2+x
iii) you can do this by integration or by normal area rules because these shapes are triangles
iiii) use the normal triangles. I will post graph

 

[jkwii]

ok cool but integration with limits is not the same as finding areas. go figure.

J.

 

[conics2008]

ok cool but integration with limits is not the same as finding areas. go figure.

J.

Integration is the process of find the area between a curve.

 

[Trebla]

Using the definition:
| x | = x, if x ≥ 0
= - x, if x < 0

∫³_-3(2 - |x|)dx = ∫³₀(2 - x)dx + ∫⁰_-3(2 + x)dx

then integrate and evaluate...

 

[Affinity]

Integration is the process of find the area between a curve.

up to a point.