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: 10

Answer

$$ - \frac{{656}}{{15}}$$

Work Step by Step

$$\eqalign{ & \int_0^4 { - \left( {3{x^{3/2}} + {x^{1/2}}} \right)} dx \cr & = - \int_0^4 {\left( {3{x^{3/2}} + {x^{1/2}}} \right)} dx \cr & {\text{integrate by using }}\int_a^b {{x^n}dx} = \left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right)_a^b \cr & = - \left( {\frac{{3{x^{5/2}}}}{{5/2}} + \frac{{{x^{3/2}}}}{{3/2}}} \right)_0^4 \cr & = - \left( {\frac{{6{x^{5/2}}}}{5} + \frac{{2{x^{3/2}}}}{3}} \right)_0^4 \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = - \left[ {\left( {\frac{{6{{\left( 4 \right)}^{5/2}}}}{5} + \frac{{2{{\left( 4 \right)}^{3/2}}}}{3}} \right) - \left( {\frac{{6{{\left( 0 \right)}^{5/2}}}}{5} + \frac{{2{{\left( 0 \right)}^{3/2}}}}{3}} \right)} \right] \cr & {\text{simplifying}} \cr & = - \left[ {\left( {\frac{{192}}{5} + \frac{{16}}{3}} \right) - \left( {0 + 0} \right)} \right] \cr & = - \left( {\frac{{656}}{{15}}} \right) \cr & = - \frac{{656}}{{15}} \cr} $$
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