Q. 41 - Calculus

Question

 Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer. 

    ${\int }_{-3}^2 \frac{2x}{x^2+5}dx$

Step-by-Step Solution

Answer

Ans:   The exact value is,  ${\int }_{-3}^2 \frac{2x}{x^2+5}dx =\ln \left(9\right)-\ln \left(14\right)$

Step 1. given information.

given,

       ${\int }_{-3}^2 \frac{2x}{x^2+5}dx$

Step 2. The objective is to determine the exact value of the definite integral.

The exact value is calculated as shown below, 

  $\begin{array}{r}{\int }_{-3}^2 \frac{2x}{x^2+5}dx\\ =2{\int }_{-3}^2 \frac{x}{x^2+5}dx\\ =2\int \frac{1}{2u}du\left[u=1+x^2,du=2xdx\right]\\ =2\left(\frac{1}{2}\right)\int \frac{1}{u}du\\ =[\ln (\left|u\right|){]}_{-3}^2\\ ={\left[\ln \left|5+x^2\right|\right]}_{-3}^2\\ =\ln \left|5+2^2\right|-\ln \left|5+(-3{)}^2\right|\\ =\ln \left(9\right)-\ln \left(14\right)\end{array}$

 Therefore, the exact value is, $\ln \left(9\right)-\ln \left(14\right)$

Step 3. Check

The required graph is,