# Question about integral and Abs function

1. Feb 8, 2010

### 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?

2. Feb 8, 2010

### 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

3. Feb 8, 2010

### njama

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.

4. Feb 8, 2010

### Mute

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.

5. Feb 9, 2010

### 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)