Question

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

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

Short answer

Ans: The exact value of, ${\int }_{-3}^2 \frac{x^2+5}{2x}dx=\frac{5\left(2\right(\ln \left(2\right)+\ln (-1))-2\ln (-3)-1)}{4}$

Step-by-step solution

Step 1. Given information.

given expression,

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

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

The exact value is calculated as shown below,

$$\begin{gathered} {\int }_{-3}^2 \frac{x^2+5}{2x}dx \\ =\frac{1}{2}{\int }_{-3}^2 xdx+\frac{5}{2}{\int }_{-3}^2 \frac{1}{x}dx \\ =[\frac{x^2}{4}+\frac{5\ln \left(x\right)}{2}{]}_{-3}^2 \\ =[\frac{5\ln \left(\right|x\left|\right)+\frac{x^2}{2}}{2}{]}_{-3}^2 \\ =\frac{5\ln \left(2\right)+5\ln (-1)-5\ln (-3)-\frac{5}{2}}{2} \\ =\frac{5\left(2\right(\ln \left(2\right)+\ln (-1))-2\ln (-3)-1)}{4} \end{gathered}$$

Therefore, the exact value is $\frac{5\left(2\right(\ln \left(2\right)+\ln (-1))-2\ln (-3)-1)}{4}$

Step 3. Check: 

The required graph is,