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: 29

Answer

$\frac{82}{5}$

Work Step by Step

Evaluate the integral: $\int^{4}_{1}\frac{2+x^2}{\sqrt x}dx$ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(x)$: $F(x)=\int\frac{2+x^2}{\sqrt x}dx$ $F(x)=\int (\frac{2}{\sqrt x} + \frac{x^2}{\sqrt x})dx$ $F(x)=\int 2x^{-1/2} + x^2(x^{-1/2})dx$ $F(x)=\int 2x^{-1/2} + x^{3/2}dx$ $F(x)=4x^{1/2} + \frac{2}{5}x^{5/2}$ $F(x)=4\sqrt x + \frac{2}{5}{\sqrt x}^5$ Now evaluate $F(b)−F(a)$: $F(4)−F(1)$ $(4\sqrt 4 + \frac{2}{5}{\sqrt 4}^5) - (4\sqrt 1 + \frac{2}{5}{\sqrt 1}^5)$ $(8+ \frac{2}{5}(32))-(4 + \frac{2}{5})$ $4 + \frac{62}{5}$ $\frac{82}{5}$
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