Question

Integration help..

Answer 1

\[\int\limits_{-1}^{2} |x+1|+|x|+|x-1|dx\]

Answer 2

@hartnn

Answer 3

idk how to break this lol..it is discontinuous at -1 0 1 so how do i put the limits? like we do -1 to 0 ,0 to 1..etc..it cant be done here?

Answer 4

yeah correct
-1 to 0 , 0 to 1 , 1 to 2
break it into 3 integras

Answer 5

\[ \int\limits_{-1}^{1}x+1-x-x+1+\int\limits_0 ^1 x+1+x+1-x+\int\limits_1^2x+1+x+x-1\]
is this okay?

Answer 6

\[\left| x+1 \right|=x+1,if x+1>0,or x>-1\]
So in the given interval x+1 is always positive.

Answer 7

\[\int\limits_{-1}^{1}2-x + \int\limits_0^1 x+2 + \int\limits_1^23x\]

Answer 8

is it okay? @hartnn

Answer 9

i think yes, you are correct

Answer 10

@surjithayer aso gets the same result as u

Answer 11

\[\left| x \right|=-x,if x<0,\left| x \right|=x,if x>0\]

Answer 12

\[\left| x-1 \right|=-\left( x-1 \right),if x<1 and \left| x-1 \right|=x-1, if x>1\]

Answer 13

Please don't forget to add dx in each integral.

Answer 14

yes

Answer 15

but m nt getting the ans :S

Answer 16

what is the answer.
I think you are doing well.

Answer 17

i got 3+9/2

Answer 18

15/2

Answer 19

oh no wait

Answer 20

for first integral i got 6

Answer 21

for 2nd i got 5/2
for 3rd i got 9/2

Answer 22

the 1st integral is -1 to 0
not -1 to 1

Answer 23

yes i know

Answer 24

0- [2(-1)-1/2] = 0 +2 +1/2
how u got 6 ?

Answer 25

5/2 sorry :|

Answer 26

5/2+7

Answer 27

19/2 thanks !

Answer 28

welcome ^_^