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: 35

Answer

$\frac{17}{2}$

Work Step by Step

Evaluate the Integral: $\int^{2}_{1}\frac{v^5+3v^6}{v^4}dv $ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(x)$: $F(x) = \int\frac{v^5 +3v^6}{v^4}dv$ $F(x) = \int \frac{v^5}{v^4}dv +\int\frac{3v^6}{v^4}dv$ $F(x) = \int{vdv} + \int3v^2dv$ $F(x) = \frac{v^2}{2} + \frac{3v^3}{3}$ $F(x) = \frac{v^2}2 + v^3 $ Now Evaluate: $F(b) - F(a)$ $F(2) - F(1)$ $(\frac{(2)^2}{2}+(2)^3) - (\frac{(1)^2}{2} + (1)^3)$ $(2 + 8) - (\frac{1}{2}+1)$ $10 - \frac{3}{2}$ $\frac{17}{2}$
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