Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Review - Exercises - Page 429: 11

Answer

$ - \frac{{13}}{6}$

Work Step by Step

$$\eqalign{ & \int_{ - 1}^0 {\left( {{x^2} + 5x} \right)} dx \cr & {\text{Integrate}}{\text{, recall that }}\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr & = \left[ {\frac{{{x^{2 + 1}}}}{{2 + 1}} + 5\left( {\frac{{{x^{1 + 1}}}}{{1 + 1}}} \right)} \right]_{ - 1}^0 \cr & = \left[ {\frac{{{x^3}}}{3} + 5\left( {\frac{{{x^2}}}{2}} \right)} \right]_{ - 1}^0 \cr & = \left[ {\frac{{{x^3}}}{3} + \frac{{5{x^2}}}{2}} \right]_{ - 1}^0 \cr & {\text{Evaluate the limits and simplify}} \cr & = \left[ {\frac{{{{\left( 0 \right)}^3}}}{3} + \frac{{5{{\left( 0 \right)}^2}}}{2}} \right] - \left[ {\frac{{{{\left( { - 1} \right)}^3}}}{3} + \frac{{5{{\left( { - 1} \right)}^2}}}{2}} \right] \cr & = \left[ 0 \right] - \left[ { - \frac{1}{3} + \frac{5}{2}} \right] \cr & = - \frac{{13}}{6} \cr} $$
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