Using antiderivative to calculate complex integral $$\int_{1}^{3}(z-2)^3 dz $$
I get the following -
$$\frac{1}{4}[(3-2)^3 - (1-2)^3] = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
However the answer sheet I have just show it reduced to $\frac{1}{4} - \frac{1}{4} = 0$
Cant see how they are getting a - instead of a +...what am I missing?

The integral of $(z-2)^3$ is $\frac14 (z-2)^4+c$ not $\frac14 (z-2)^3+c$, so you should get $$\tfrac{1}{4}(3-2)^4 - \tfrac{1}{4}(1-2)^4 =0.$$
