Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises - Page 376: 87

Answer

$$\frac{2}{3}$$

Work Step by Step

\eqalign{ & \int_1^4 {\frac{{x - 2}}{{\sqrt x }}dx} \cr & {\text{split the numerator}} \cr & = \int_1^4 {\left( {\frac{x}{{\sqrt x }} - \frac{2}{{\sqrt x }}} \right)} dt \cr & {\text{writting }}\sqrt x {\text{ as }}{x^{1/2}} \cr & = \int_1^4 {\left( {\frac{x}{{{x^{1/2}}}} - \frac{2}{{{x^{1/2}}}}} \right)} dt \cr & = \int_1^4 {\left( {{x^{1/2}} - 2{x^{ - 1/2}}} \right)} dt \cr & {\text{integrate using the power rule for integration}} \cr & = \left. {\left( {\frac{{{x^{3/2}}}}{{3/2}} - 2\left( {\frac{{{x^{1/2}}}}{{1/2}}} \right)} \right)} \right|_1^4 \cr & = \left. {\left( {\frac{{2{x^{3/2}}}}{3} - 4{x^{1/2}}} \right)} \right|_1^4 \cr & {\text{using The Fundamental Theorem}} \cr & = \left( {\frac{{2{{\left( 4 \right)}^{3/2}}}}{3} - 4{{\left( 4 \right)}^{1/2}}} \right) - \left( {\frac{{2{{\left( 1 \right)}^{3/2}}}}{3} - 4{{\left( 1 \right)}^{1/2}}} \right) \cr & {\text{simplify}} \cr & = \left( {\frac{{16}}{3} - 8} \right) - \left( {\frac{2}{3} - 4} \right) \cr & = - \frac{8}{3} + \frac{{10}}{3} \cr & = \frac{2}{3} \cr}

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