Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.4 Indefinite Integrals and the Net Change Theorem - 4.4 Exercises - Page 337: 24

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.
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