Integrating $\int_{0}^{2} (1-x)^2 dx$ I solved this  integral
$$\int_{0}^{2} (1-x)^2 dx$$
by operating the squared binomial, first. 
But, I found in some book, that it arrives at the same solution and I don't understand why it appears a negative simbol. This is the author solution:
$$\int_{0}^{2} (1-x)^2 dx = -\frac{1}{3}(-x+1)^3|_{0}^{2}=\frac{2}{3}$$

There are two places where you need to be careful about the sign. The first place is when you make the derivative of the substitution (if your substitution is 1-x = t, then after making the derivative you get -dx = dt). The second place is when you change the limits of the integration due to the substitution. The lower limit becomes 1 and the upper limit becomes -1. You can find a complete solution to this problem here
http://www.e-academia.cz/solved-math-problems/definite-integral-with-substitution.php
