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 gopositiveagain, with values that are 'opposite' to thenegativevalues of (x - 2). So the graph of y = x|x - 2| is thereflectionof 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
positivenumber, because (as we've just explained) the graph isabovethe x-axis.- Add the two areas together.

OK?

Grandad