Calculus 8th Edition

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

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises: 27

Answer

$-\frac{37}{6}$

Work Step by Step

Evaluate the integral: $\int^{1}_{0}(u+2)(u-3) du$ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(x)$: $F(x)=\int(u+2)(u-3) du$ $F(x)=\int(u^2 - u -6)dx$ $F(x)=\frac{u^3}{3} -\frac{u^2}{2} - 6u$ Now evaluate $F(b)−F(a)$: $F(1)−F(0)$ $(\frac{(1)^3}{3} -\frac{(1)^2}{2} - 6(1)) - (\frac{(0)^3}{3} -\frac{(0)^2}{2} - 6(0))$ $\frac{1}{3} - \frac{1}{2} -6 -0$ $\frac{2}{6}- \frac{3}{6} - \frac{36}{6}$ $-\frac{37}{6}$
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