Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 50

Answer

=$\frac{7}{2}$

Work Step by Step

$\int^0_1(3x^2+x-5) \ dx$=$-\int^1_0(3x^2+x-5) \ dx$ =$-[3\int^1_0x^2 \ dx+\int^1_0x \ dx-\int^1_05 \ dx$ =$-[3(\frac{1^3}{3}-\frac{0^3}{3})+(\frac{1^2}{2}-\frac{0^2}{2})-5(1-0)]$ =$-(\frac{3}{2}-5)$=$\frac{7}{2}$
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