# Thread: Integration with absolute values?

1. ## Integration with absolute values?

Hey guys I'm stuck on this problem, and it's likely to be on my final tomorrow.

Compute the area bounded by the graph y = x |x − 2|, the x-axis, the
y-axis and the vertical line x = 3.

I'm not sure how to go about integrating with the absolute value there, can anyone please help me out?

Thanks in advance!

2. ## Integration with absolute values

Hi -

We need to think about the graphs of y = x|x - 2| and y = x(x - 2) between x = 0 (the y-axis) and x = 3. So here goes...

The only time |x - 2| and (x - 2) are different is when (x - 2) goes negative: when that happens, |x - 2| stays positive but has the same numerical value as (x - 2).

So, if x > 2, (x - 2) and |x - 2| are both the same, and the graph of y = x|x - 2| is the same as the graph of y = x(x - 2), and lies above the x-axis.

When x = 2 and x = 0, both x(x - 2) and x|x - 2| have the value zero, of course. So each graph crosses the x-axis at these points.

What happens when x lies between 0 and 2? Well, (x - 2) goes negative, of course. So the graph of y = x(x - 2) will go below the x-axis, because the value of y will be negative: a positive times a negative.

But when x lies between 0 and 2, |x - 2| will go positive again, with values that are 'opposite' to the negative values of (x - 2). So the graph of y = x|x - 2| is the reflection of y = x(x - 2) in the x-axis, for values of x between 0 and 2.

So, here's how you solve your integration:

• Using the above information, sketch the graph of y = x|x - 2| between x = 0 and x = 3.
• Work out the area in two parts: (1) between x = 0 and x = 2; and (2) between x = 2 and x = 3. Each area will be a positive number, because (as we've just explained) the graph is above the x-axis.
• Add the two areas together.

OK?
Grandad

3. Okay, I've got it.

Thanks a lot!