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: 33

Answer

$\displaystyle\int\limits_0^1(1+r)^{3}dr=\dfrac{15}{4}$

Work Step by Step

$\displaystyle\int\limits_0^1(1+r)^{3}dr$ Let's apply the cube of a binomial rule to the integral. This rule is $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{2}$ $\displaystyle\int\limits_0^1(1+r)^{3}dr=\int\limits_0^1(1+3r^{2}+3r+r^{3})dr=...$ Now, integrate each term separately and apply the second part of the fundamental theorem of calculus: $...=\displaystyle\int\limits_0^1dr+3\int\limits_0^1r^{2}dr+3\int\limits_0^1rdr+\int\limits_0^1r^{3}dr=...$ $...=r+(3)\Big(\dfrac{1}{3}\Big)r^{3}+(3)\Big(\dfrac{1}{2}\Big)r^{2}+\dfrac{1}{4}r^{4}\Big|_0^1=r+r^{3}+\dfrac{3}{2}r^{2}+\dfrac{1}{4}r^{4}\Big|_0^1$ $...=\Big[1+(1)^{3}+\dfrac{3}{2}(1)^{2}+\dfrac{1}{4}(1)^{4}\Big]-\Big[0+(0)^{3}+\dfrac{3}{2}(0)^{2}+\dfrac{1}{4}(0)^{4}\Big]$ $...=1+1+\dfrac{3}{2}+\dfrac{1}{4}=\dfrac{15}{4}$
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