Okay looking at the diagram below, notice the red line. It is the shape of a
trapezium.You can find the area under the
curve and from this value MINUS the area under the
line.
First I found the two points at which the curve and line intersect :
x
2 - 4x +4 = 7-2x
Hence, x
2-2x -3 =0
Solving for x gives : 3 and -1
Inputting the above x values into 7-2x = y give corresponding y coordinates of : 1 and 9
Finding the area under the line :
*4=20)
Next I integrate the curve to get :
^3}{3})
Inputting 3 into above gives a value of 1/3 .
Inputting -1 into abov gives : -9
Hence 1/3 - - 9 =

Hence, 20 -

=
