Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 19


$\int\limits_1^3 (x^{2}+2x-4) dx=\dfrac{26}{3}$

Work Step by Step

$\int\limits_1^3 (x^{2}+2x-4) dx$ Integrate each term: $\int\limits_1^3 x^{2} dx +\int\limits_1^3 2x dx-\int\limits_1^3 4 dx=...$ Take the constants out of the integral and evaluate each individual integral: $...=\int\limits_1^3 x^{2} dx+2\int\limits_1^3 x dx-4\int_\limits1^3dx=...$ $...=\dfrac{1}{3}x^{3}+(2)(\dfrac{1}{2})x^{2}-4x\Big|_1^3=...$ $...=\dfrac{1}{3}x^{3}+x^{2}-4x\Big|_1^3=...$ Apply the second part of the fundamental theorem of calculus to get the result: $...=\Big[\dfrac{1}{3}(3)^{3}+(3)^{2}-4(3)\Big]-\Big[\dfrac{1}{3}(1)^3+(1)^2-4(1)\Big]=...$ $...=\dfrac{1}{3}(27)+9-12-\dfrac{1}{3}-1+4=...$ $...=9+9-12-\dfrac{1}{3}-1+4=\dfrac{26}{3}$
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