Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Solve given definite integrals.
$\int_{0}^{4} x \sqrt{x^{2} + 4} d x$

$\frac{40 \sqrt{5}}{3} - \frac{8}{3} .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1. Given Information

The integral is as follows.

$\int_{0}^{4} x \sqrt{x^{2} + 4} d x$

The objective is to solve the integral.

Step2. Assumption

The integral is solved below.

$\int x \sqrt{x^{2} + 4} d x$

$= \int \frac{\sqrt{u}}{2} d u [u = x^{2} + 4, d u = 2 x d x]$

$\begin{aligned} = \frac{1}{2} \int \sqrt{u} d u \\ = \frac{u^{\frac{3}{2}}}{3} \\ = \frac{{(x^{2} + 4)}^{\frac{3}{2}}}{3} \\ = \frac{1}{3} {(x^{2} + 4)}^{\frac{3}{2}} \end{aligned}$

Step3. Solution

The definite integral is solved below.

$\begin{aligned} \int_{0}^{4} x \sqrt{x^{2} + 4} d x \\ = {[\frac{1}{3} {(x^{2} + 4)}^{\frac{3}{2}}]}_{0}^{4} \\ = [\frac{1}{3} {(4^{2} + 4)}^{\frac{3}{2}}] - [\frac{1}{3} {(0^{2} + 4)}^{\frac{3}{2}}] \\ = \frac{40 \sqrt{5}}{3} - \frac{8}{3} \end{aligned}$

Therefore, the value is $\frac{40 \sqrt{5}}{3} - \frac{8}{3} .$

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Calculus

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Short Answer

Solve given definite integrals.
$\int_{0}^{4} x \sqrt{x^{2} + 4} d x$

Step by Step Solution

Step1. Given Information

Step2. Assumption

Step3. Solution

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Solve given definite integrals.
$\int_{0}^{4} x \sqrt{x^{2} + 4} d x$