Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 38

Answer

$$0$$

Work Step by Step

$$\eqalign{ & \int_0^4 {x\left( {x - 2} \right)\left( {x - 4} \right)} dx \cr & {\text{multiply the integrand}} \cr & = \int_0^4 {x\left( {{x^2} - 6x + 8} \right)} dx \cr & = \int_0^4 {\left( {{x^3} - 6{x^2} + 8x} \right)} dx \cr & {\text{using power rule for integration}} \cr & = \left. {\left( {\frac{{{x^4}}}{4} - 6\left( {\frac{{{x^3}}}{3}} \right) + 8\left( {\frac{{{x^2}}}{2}} \right)} \right)} \right|_0^4 \cr & = \left. {\left( {\frac{{{x^4}}}{4} - 2{x^3} + 4{x^2}} \right)} \right|_0^4 \cr & {\text{Using The Fundamental Theorem}} \cr & = \left( {\frac{{{{\left( 4 \right)}^4}}}{4} - 2{{\left( 4 \right)}^3} + 4{{\left( 4 \right)}^2}} \right) - \left( {\frac{{{{\left( 0 \right)}^4}}}{4} - 2{{\left( 0 \right)}^3} + 4{{\left( 0 \right)}^2}} \right) \cr & {\text{simplify}} \cr & = \left( {64 - 128 + 64} \right) - \left( {0 - 0 + 0} \right) \cr & = 0 \cr} $$
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