Integrating the curve y=3x-x
2 from the x-coordinate of A (-2) to the origin (0) will give you the area bounded between the curve and the
x-axis. The same applies when you integrate the line y=x-8. When there are segments of area above and below the x-axis, you will not be able to just integrate the entire segment as the areas below the x-axis will be calculated as
negative values which will cancel out the areas above the x-axis. This is a good example illustrating how you can get a 'zero area' answer after integration:
http://revisionmaths.com/advanced-level-maths-revision/pure-maths/calculus/area-under-curveI'll assume you know how to find the coordinates of A and B and that the problem lies in the second part of the question. To get the shaded area, you'll need to break the integration process into three regions:
A) From x=-2 to x=0
1. Integrate y=x-8 to find the area bounded between the line and the x-axis. You'll notice that it's a negative answer because the area is
underneath the x-axis.
2. Integrate y=3x-x
2 to find the area bounded between the curve and the x-axis.
3. The shaded area will be the difference between the absolute value of the answer in 1. and the absolute value of the answer in 2. (Absolute value basically means ignore the negative sign and just look at the 'magnitude' of the area)
B) From x=0 to x=3 (from the origin to the second intersection between the curve and the x-axis)
1. Integrate y=x-8 to find the area bounded between the line and the x-axis.
2. Integrate y=3x-x
2 to find the area bounded between the curve and the x-axis.
3. The shaded area will be the sum of the absolute value of the answer in 1. and the value of the answer in 2.
C) From x=3 to x=4 (from the second intersection between the curve and the x-axis to the x-coordinate of B)
I'll let you figure this one out by yourself. It's almost the same as part A.