I How do you solve the definite integral ∫|x^3-x|dx with limits from -1 to 2?

[Prasun-rick]

∫|x^3-x|dx with limits from -1 to 2.Can anyone kindly show the step wise solution or suggest how to proceed .Thanks in advance​

 

[S.G. Janssens]

Can you tell us first what makes it troubling for you?

 

[BvU]

First step: make a drawing !

 

[Prasun-rick]

Can you tell us first what makes it troubling for you?

The modulus sign! I In which limit I have to take positive and negative in the modulus I could not understand!

 

[Prasun-rick]

First step: make a drawing !

Well I am not able to draw precisely and that is where the problem lies !

 

[BvU]

Doesn't have to be precise. Simply investigate the function a little.
Where it comes from on the left, where it goes through zero and where it goes to on the right.
Perhaps there is even a little symmetry to explore and make good use of ...

 

[Prasun-rick]

Doesn't have to be precise. Simply investigate the function a little.
Where it comes from on the left, where it goes through zero and where it goes to on the right.
Perhaps there is even a little symmetry to explore and make good use of ...

well kindly please tell me the terminal points ! I found it as -1,0 and 1 am I correct??

 

[BvU]

I don't know what terminal points are but the zero points of $\ \ x^3-x = x(x+1)(x-1) \ \$ are indeed -1, 0, 1 .

Between -1 and 0 two out of three factors are < 0 and the other is > 0, so the expression is positive.
Between 0 and 1 ...
There is antisymmetry : $\ \ (-x)^3 -(-x) = -(x^3 - x)$, so ...

 

[Prasun-rick]

I don't know what terminal points are but the zero points of $\ \ x^3-x = x(x+1)(x-1) \ \$ are indeed -1, 0, 1 .

Between -1 and 0 two out of three factors are < 0 and the other is > 0, so the expression is positive.
Between 0 and 1 ...
There is antisymmetry : $\ \ (-x)^3 -(-x) = -(x^3 - x)$, so ...

Thanks I finally got that !Thank you very much !

 

[chiro]

Hey Prasun-rick.

You will have to split up the function within the absolute value function so that it is positive and negative and create separate integrals for each section.

Just remember that you define |x| = x if x > 0 and -x if x < 0 (and 0 if x = 0).