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Q. 79

Expert-verified
Found in: Page 465

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Solve given definite integrals.${\int }_{0}^{4}x\sqrt{{x}^{2}+4}dx$

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

See the step by step solution

## Step1. Given Information

The integral is as follows.

${\int }_{0}^{4} x\sqrt{{x}^{2}+4}dx$

The objective is to solve the integral.

## Step2. Assumption

The integral is solved below.

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

$=\int \frac{\sqrt{u}}{2}du\phantom{\rule{1em}{0ex}}\left[u={x}^{2}+4,du=2xdx\right]$

$\begin{array}{r}=\frac{1}{2}\int \sqrt{u}du\\ =\frac{{u}^{\frac{3}{2}}}{3}\\ =\frac{{\left({x}^{2}+4\right)}^{\frac{3}{2}}}{3}\\ =\frac{1}{3}{\left({x}^{2}+4\right)}^{\frac{3}{2}}\end{array}$

## Step3. Solution

The definite integral is solved below.

$\begin{array}{r}{\int }_{0}^{4} x\sqrt{{x}^{2}+4}dx\\ ={\left[\frac{1}{3}{\left({x}^{2}+4\right)}^{\frac{3}{2}}\right]}_{0}^{4}\\ =\left[\frac{1}{3}{\left({4}^{2}+4\right)}^{\frac{3}{2}}\right]-\left[\frac{1}{3}{\left({0}^{2}+4\right)}^{\frac{3}{2}}\right]\\ =\frac{40\sqrt{5}}{3}-\frac{8}{3}\end{array}$

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