## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises: 15

#### Answer

$G(t) = 2t^{\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + \frac{2}{5}t^{\frac{5}{2}} + C$

#### Work Step by Step

$g(t) = \frac{1 + t + t^{2}}{\sqrt t}$ Expand the function: $g(t) = \frac{1}{\sqrt t} + \frac{t}{\sqrt t} + \frac{t^{2}}{\sqrt t}$ Convert the square roots into exponents: $g(t) = \frac{1}{t^{\frac{1}{2}}} + \frac{t}{t^{\frac{1}{2}}} + \frac{t^{2}}{t^{\frac{1}{2}}}$ Simplify: $g(t) = t^{\frac{-1}{2}} + t^{\frac{1}{2}} + t^{\frac{3}{2}}$ Apply the Power rule: $\frac{x^{a+1}}{a+1}$: $G(t) = \frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1} + \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1} + \frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}$ $G(t) = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + \frac{t^{\frac{5}{2}}}{\frac{5}{2}}$ $G(t) = 2t^{\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + \frac{2}{5}t^{\frac{5}{2}}$ Add the constant to the solution: $G(t) = 2t^{\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + \frac{2}{5}t^{\frac{5}{2}} + C$

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