Question

Compute integral of discontinuous function

Answer 1

what is sign mean

Answer 2

sign function or signum function
http://en.wikipedia.org/wiki/Sign_function

Answer 3

so when is x-x^3 positive ?
and when it is negative ?

Answer 4

when x>1 (x-x^3) is negative, if x<-1 (x-x^3) is positive? T^T im confused

Answer 5

when x>1 (x-x^3) is negative , correct!
how x<-1 ?
did u mean x<1 ?
if not, when x<1, is x-x^3 positive or negative ?

Answer 6

say for example x=1/2
1/2 - 1/8, positive or negative ?

Answer 7

x<1 the func is 0 right

Answer 8

why ?

Answer 9

oh no sry, wrong haha i mean, positive?

Answer 10

okay idk why do we need to do that..

Answer 11

then yes,
this is the way to solve it algebraically.
x-x^3 is positive when
x-x^3 > 0
x> x^3
1 >x^2 (since the limits indicate x is positive from 0 to 3, we will consider this range only)
so, 1 > x
so, x>1 function is positive
x<1 function is negative

Answer 12

did you read the definition of signum function ?

Answer 13

its something that extracts the sign of the real number? why do u need to extract it?

Answer 14

sgn x =1 when x> 0 , that is when x is positive
so we need the value of x when your function is positive

Answer 15

so sign (x-x^3) = 1 when x-x^3 > 0
that is when x>1
got this ?

Answer 16

^^^ x < 1 ??

Answer 17

0<x<1 positive
1<x<3 negative

Answer 18

yes, sorry for the typo

sign (x-x^3) = -1 when x-x^3 < 0
that is when x>1
got this ?

Answer 19

ah! okay, then so nwo that weve got that x>1 func is positive, x<1 is negative, then

Answer 20

the other way around.

Answer 21

we split the integral, from 0 to 1 and 1 to 3

Answer 22

careful! :P

Answer 23

from 0 to 1, x-x^3 >0 , so sign (x-x^3) = +1

Answer 24

oh ya, hm ok so,\[\int\limits_{0}^{1}sign(x-x^3)dx+\int\limits_{1}^{3}sign(x-x^3)dx\]

Answer 25

from 1 to 3, x-x^3 <0 , so sign (x-x^3) = -1

Answer 26

\(\int\limits_{0}^{1}1dx+\int\limits_{1}^{3}(-1)dx\)

Answer 27

oh, so u gotta follow the sign rule we've come up just now..

Answer 28

1-2=-1?

Answer 29

yes

Answer 30

You've been my hero for a long time hartnn. Nothing seems to stump you ;)

Answer 31

lol, just when you said that, i was thinking, how the hell am i gonna do the 2nd one :P
i have no idea about it :O

Answer 32

i guess splitting from 0 to 1 and 1 to 2 will work...

Answer 33

Isn't it just e^x evaluated from 0 to 2? Are those special symbols or something?

Answer 34

We just need to find when e^x is 1 and 2 first

Answer 35

yeah, that |_ ... _| means the floor function

Answer 36

oh man... darn little symbols

Answer 37

yes, what ranga said
when is e^x = 1
when is e^x = 2 ?

Answer 38

hm, e^x=1 when x=0?

Answer 39

e^x=2 when x=ln2? haha idk

Answer 40

e^x = 2 when x = ln(2)
e^x = 3 when x = ln(3)
....

Answer 41

yeah, from 0 to ln2

Answer 42

http://www.wolframalpha.com/input/?i=y%3D+e%5Ex+from+x%3D0+to+x%3D2

you know about floor functions ?

Answer 43

now that you said it... rings a bell

Answer 44

http://www.mathsisfun.com/sets/function-floor-ceiling.html

Answer 45

yup, its the awkward kind of function like staircase haha

Answer 46

oh my, this doesn't hit 2 until between ln7 and ln8... does that mean multiple integrals?

Answer 47

yes

Answer 48

wait wanna vomit blood.. lol!

Answer 49

lol

Answer 50

so we split the range 0 to 2 into many sub-intervals,
like
from 0(which is ln 1) to ln 2, then floor(e^x)= ... ?
from ln 2 to ln 3 , floor(e^x) =... ?
.....
from ln 7 to 2, floor (e^x) = ... ?

Answer 51

yeah... I like it

Answer 52

well "like" is subjective

Answer 53

haha kidding :p okay 1st one e^x=1, ln2 to ln 3 e^x=2......., 3, 4, 5, 6 ,7?

Answer 54

correct!
just plug in values!

Answer 55

okay i think i got it then! thanks! :D

Answer 56

the 3rd one too ?

Answer 57

3rd one is same as the 1st one right? i mean also got the sign

Answer 58

yes, same 'type'

Answer 59

just need to find when cos x is positive and negative in the range 0 to pi and then split the intrgral...

Answer 60

at x=pi func is negative, at x=0 func is positive

Answer 61

we'll need "interval"

Answer 62

cos(x) is positive for 0 < x < pi/2
cos(x) is negative for pi/2 < x < pi

Answer 63

from 0 to which value is cos x positive ?

Answer 64

yes, that

Answer 65

so you'll split your integral from 0 to pi/2
and pi/2 to pi

Answer 66

i see, okay, if thats the case, \[\int\limits_{0}^{\frac{ \pi }{ 2 }}x(1)dx+\int\limits_{-\frac{ \pi }{ 2 }}^{\pi}x(-1)d\]

Answer 67

*dx

Answer 68

lower limit on second integral should be +pi/2

Answer 69

oh ya, omg

Answer 70

everything else was correct

Answer 71

okay great! thanks alot guys!! i appreciate! :D

Answer 72

thank you too for asking good questions! :)