Dear John Brindley,
the integral from a to b is not the area, but the signed area.
integral -cos(x) from 0 to pi/2 = -1
integral -cos(x) from pi/2 to pi = 1
intergal from 0 to pi = (-1) + (+1) = 0.
The integral of -cosx is sinx. If I wish to define the interval between zero and pi to calculate the area of f(x) above and below the x-axis, I get:
0 - 0 = 0.
If I redefine the interval as 1/2pi to 3/2pi, I get 2 square units.
But shifting the same interval up and down should not make any difference to the area of f(x) - but it does seem to!
Can somebody tell me where I'm going wrong?