Answer
$ -\frac{4}{3} $
Work Step by Step
We must simplify the integrand before evaluating it. First, we expand the term $(1 - t)^2 $ and get $ 1 - 2t + t ^2$. Then we multiply the resulting polynomial by $t$ and get $ t - 2t^2 +t^3$.
We can now evaluate the integral of this simplified polynomial, which is $\frac{t^2}{2} - \frac{2t^3}{3} + \frac{t^4}{4} $
We now substitute the two bounds of integration, $1$ and $-1$ into $t$ and get the difference
($\frac{1^2}{2} - \frac{2(1)^3}{3} + \frac{1(1)^4}{4} $) - ($\frac{(-1)^2}{2} - \frac{2(-1)^3}{3} + \frac{(-1)^4}{4} $) = ($\frac{1}{2} - \frac{2}{3} + \frac{1}{4})$ - $(\frac{1}{2} + \frac{2(-1)^3}{3} + \frac{(-1)^4}{4} $) $ = -\frac{4}{3} $
Note that the main difference between substituting $1$ into $t$ compared to substituting $-1$ into $t$ is that the fraction$ \frac{2(-1)^3}{3}$ is added instead of subtracted because the $-1$ remains negative due to the odd exponent.