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: 19

Answer

$\frac{26}{3}$

Work Step by Step

Evaluate the integral: $\int^3_1 (x^2 +2x -4) dx$ Recall the 2nd part of the Fundamental Theorem of Calculus: $\int_a^bf(x)dx = F(b) - F(a)$ Find $F(x)$: $F(x) = \int (x^2 +2x-4)dx $ $F(x) = \frac{x^3}{3} + \frac{2x^2}{2} -\frac{4x^1}{1} $ $F(x) =\frac{x^3}{3}+x^2 -4x $ Now evaluate $F(b) - F(a)$: $F(3) - F(1)$ $(\frac{(3)^3}{3}+(3)^2 -4(3) )- (\frac{(1)^3}{3}+(1)^2 -4(1)) $ $(9+9 -12) - (\frac{1}{3} +1 - 4)$ $6 + 3 - \frac{1}{3}$ $\frac{27}{3} -\frac{1}{3} = \frac{26}{3}$
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