Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.4 The Fundamental Theorem of Calculus - 7.4 Exercises - Page 395: 9

Answer

$$ - \frac{{16}}{3}$$

Work Step by Step

$$\eqalign{ & \int_0^4 {2\left( {{t^{1/2}} - t} \right)} dt \cr & = \int_0^4 {\left( {2{t^{1/2}} - 2t} \right)} dt \cr & {\text{use }}\int {{t^n}dt = \frac{{{t^{n + 1}}}}{{n + 1}} + C,{\text{ so}}} \cr & = \left( {2\left( {\frac{{{t^{1/2 + 1}}}}{{1/2 + 1}}} \right) - 2\left( {\frac{{{t^2}}}{2}} \right)} \right)_0^4 \cr & = \left( {2\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right) - {t^2}} \right)_0^4 \cr & = \left( {\frac{4}{3}{t^{3/2}} - {t^2}} \right)_0^4 \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = \left( {\frac{4}{3}{{\left( 4 \right)}^{3/2}} - {{\left( 4 \right)}^2}} \right) - \left( {\frac{4}{3}{{\left( 0 \right)}^{3/2}} - {{\left( 0 \right)}^2}} \right) \cr & {\text{simplifying}} \cr & = \left( {\frac{{32}}{3} - 16} \right) - \left( 0 \right) \cr & = - \frac{{16}}{3} \cr} $$
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