## Calculus 8th Edition

$-\frac{4}{3}$
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.