Question

Another anti-derivative question!

Answer 1

This might sound stupid but how do you take the anti derivative of \[\left| x \right|\]

Answer 2

https://answers.search.yahoo.com/search;_ylt=A0LEViw.yOZU1XkAvwonnIlQ?p=how%20do%20you%20take%20the%20anti%20derivative%20of%20 |x|&.sep=&fr=yhs-mozilla-001

Answer 3

I cant find it on yahoo :/

Answer 4

When there are two lines next to x. That means x is to the power of 2 and then square rooted.

Answer 5

Answer is x

Answer 6

Oh.. I meant absolute value

Answer 7

I do not know exactly what are you talking about, but I learned that in Chemistry. I hope I helped.

Answer 8

Like the anti derivative of abs(x)

Answer 9

|x|=x if x>0
|x|=-x if x<0

Answer 10

\[\int\limits_{}^{}|x|dx \\ \text{ assuming } x \text{ is positive you have } \int\limits |x| dx=\int\limits x dx\]

Answer 11

\[\text{ assuming } x \text{ is negative you have } \int\limits |x| dx=\int\limits -x dx\]

Answer 12

there are two situations... as freckles pointed out and then you can take antiderivatives add 1 to the exponent and divide by the new exponent.

Answer 13

hmm okay, maybe asking you the entire question might be a better idea.
im finding the area between curves.
y= \[4xe ^{-x ^{2}}\] and y = abs(x)

Answer 14

the intersection points are 0 and 1.18... so I would integrate from 0 to 1.18.. (4xe^(-x^2
)) - (abs (x)) dx

Answer 15

so from there I have to find the anti derivative of them and I have no idea what to write for the abs(X) part.. Am I making any sense? :(

Answer 16

oh that was a while back.... errrr.... well you could come up with two equations one with x and one with -x -> (x) due to the sign distribution because that's what the absolute value rule is... \[\mid x \mid\] we either have
\[\mid -x \mid \rightarrow x \]
or
\[\mid x \mid \rightarrow x \]

Answer 17

oh..I guess you have one equation after all

Answer 18

why do I get the feeling that you need integration by parts :/

Answer 19

well from x=0 to x=1.18 we only have positive x

Answer 20

Would the absolute value not matter since the intersection points occur in the first quadrant?

Answer 21

|x|=x then

Answer 22

yeah what freckles said

Answer 23

ahh okay, maybe I was over thinking things haha, thank you both!!!! :D

Answer 24

np :)

Answer 25

Oh by the way, does anyone know why when i graph cos(x) in the3 calculator, it gives me a stright line?

Answer 26

i have the nspire