Question about integral and Abs function

[njama]

Hello!

I want to find

\int_{0}^{2}{|2x-3|}

I know the method for separating the integral on two integrals, but I want to try second method.

I know that

\int_{0}^{2}{|2x-3|}=\frac{|2x-3|(x-3)x}{2x-3}|_{0}^{2}

But in this case I got F(2)-F(0)=-2 and on the first method 5/2.

Why the second method does not work?

Thanks in advance.

 

[CompuChip]

What makes you think that the last equation is true?
I don't see where the x(x - 3) comes from, and how taking the derivative of sgn(2x-3) x (x - 3) is |2x - 3| :S

 

[njama]

What makes you think that the last equation is true?
I don't see where the x(x - 3) comes from, and how taking the derivative of sgn(2x-3) x (x - 3) is |2x - 3| :S

It is.

Let |2x-3|=\sqrt{(2x-3)^2}

\frac{\sqrt{(2x-3)^2}(x-3)x}{2x-3}

Now taking the derivative:

\frac{(\sqrt{(2x-3)^2}(x-3)x)'*(2x-3) - \sqrt{(2x-3)^2}(x-3)x*(2x-3)'}}{(2x-3)^2}

Also you can check it by first

x>= 3/2

(x-3)x=x^2 -3x

derivative:

2x-3

Now

x< 3/2

-(x-3)x=-x^2+3x

derivative:

-2x+3=-(2x-3)

This is a lot work to do here, so I check it again with Mathematica 5.0 and it is true. But still can't see the problem.

 

[Mute]

Hello!

I want to find

\int_{0}^{2}{|2x-3|}

I know the method for separating the integral on two integrals, but I want to try second method.

I know that

\int_{0}^{2}{|2x-3|}=\frac{|2x-3|(x-3)x}{2x-3}|_{0}^{2}

But in this case I got F(2)-F(0)=-2 and on the first method 5/2.

Why the second method does not work?

Thanks in advance.

You can't ignore the fact that there is a discontinuity at x = 3/2. The proper computation is

\left.\mbox{sgn}(2x-3)(x-3)x\right|^{2}_{3/2^+} + \left.\mbox{sgn}(2x-3)(x-3)x\right|^{3/2^-}_{0},
where 3/2^+ means infinitesimally above 3/2 and 3/2^- means infinitesimally below 3/2, such that in the first term sgn(2*x-3) = 1, while in the second it is -1. You then get

-2 - (-3/2)(3/2) - ((-3/2)(3/2) - 0) = 5/2, which is your desired result.

 

[njama]

Thanks for the reply. I think I understand.

A_1(x)=(x-3)x when x>=3/2

and

A_2(x)=-x(x-3) when x<3/2

Where A₁(x) and A₂(x) represent the area of the interval [0,x]

But if I want to find the area on the interval [0,2]

Now I find the area of [0,3/2] + (area [0,2] - area [0, 3/2])

First I must find A₂(3/2)+(A₁(2) - A₂(3/2))=A₁(2)