Chapter 6.2, Problem 33E

To determine

To evaluate integral

  ${\displaystyle \underset{0}{\overset{5}{\int }}\frac{x}{x^2+4}}dx$

Answer to Problem 33E

  ${\displaystyle \underset{0}{\overset{5}{\int }}\frac{x}{x^2+4}}dx\approx 0.9905$

Explanation of Solution

Given:Definite integral is

  ${\displaystyle \underset{0}{\overset{5}{\int }}\frac{x}{x^2+4}}dx$

Calculation: Consider the integral

  $I={\displaystyle \underset{0}{\overset{5}{\int }}\frac{x}{x^2+4}}dx$

Substitute

  $\begin{array}{l}{x}^2+4=u\\ \frac{du}{dx}=\frac{d}{dx}\left(x^2+4\right)\\ =2x\\ xdx=\frac{du}{2}\\ x=0\Rightarrow u=4\\ x=5\Rightarrow u=29\end{array}$

Now

  $\begin{array}{l}I=\frac{1}{2}{\displaystyle \underset{4}{\overset{29}{\int }}\frac{1}{u}du}\\ =\frac{1}{2}{\left[\ln \left|u\right|\right]}_4^{29}\\ =\frac{1}{2}\left[\ln 29-ln4\right]\\ \approx \frac{1}{2}\left[3.36729-1.38629\right]\\ \approx \frac{1}{2}\times 1.981\\ \approx 0.9905\end{array}$