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: 34

Answer

$\frac{17}{6}$

Work Step by Step

Evaluate the Integral: $\int^{2}_{1}\frac{s^4+1}{s^2}ds $ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(x)$: $F(x) = \int\frac{s^4+1}{s^2}ds$ $F(x) = \int\frac{s^4}{s^2}ds + \int\frac{1}{s^2}ds$ $F(x) = \int s^2ds + \int s^{-2}ds$ $F(x) = \frac{s^3}{3} - \frac{1}{s}$ $F(x) = \frac{s^4 - 3}{3s} $ Now Evaluate: $F(b) - F(a)$ $F(2) - F(1)$ $\frac{(2)^4 - 3}{3(2)} - \frac{(1)^4 -3}{3(1)}$ $\frac{16-3}{6} - \frac{1 - 3}{3}$ $\frac{13}{6} + \frac{2}{3}$ $\frac{17}{6}$
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