## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 29

#### Answer

$\displaystyle\int\limits_1^4\dfrac{2+x^{2}}{\sqrt{x}}dx=\dfrac{82}{5}$

#### Work Step by Step

$\displaystyle\int\limits_1^4\dfrac{2+x^{2}}{\sqrt{x}}dx$ Rewrite the integrand like this: $\displaystyle\int\limits_1^4\dfrac{2+x^{2}}{\sqrt{x}}dx=\int\limits_1^4\dfrac{2+x^{2}}{x^{1/2}}dx=\int\limits_1^4(2+x^{2})(x^{-1/2})dx=...$ Evaluate the product inside the integral: $...=\displaystyle\int\limits_1^4(2x^{-1/2}+x^{3/2})dx=...$ Integrate each term separately and apply the second part of the fundamental theorem of calculus: $...=\displaystyle2\int\limits_1^4x^{-1/2}dx+\int\limits_1^4x^{3/2}dx=2\Big(\dfrac{1}{1-\frac{1}{2}}\Big)x^{1-\frac{1}{2}}+\dfrac{1}{1+\frac{3}{2}}x^{1+\frac{3}{2}}\Big|_1^4$ $...=(2)\Big(\dfrac{1}{1/2}\Big)x^{1/2}+\dfrac{1}{5/2}x^{5/2}\Big|_1^4=4x^{1/2}+\dfrac{2}{5}x^{5/2}\Big|_1^4=...$ $...=4\sqrt{x}+\dfrac{2}{5}\sqrt{x^{5}}\Big|_1^4=\Big[4\sqrt{4}+\dfrac{2}{5}\sqrt{4^{5}}\Big]-\Big[4\sqrt{1}+\dfrac{2}{5}\sqrt{1^{5}}\Big]=...$ $...=4(2)+\dfrac{2}{5}\sqrt{1024}-4(1)-\dfrac{2}{5}(1)=8+\dfrac{2}{5}(32)-4-\dfrac{2}{5}=\dfrac{82}{5}$

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