Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

${\int }_1^{e-3}\frac{3\mathrm{\pi }}{2x+6}dx$

Short answer

${\int }_1^{e-3}\frac{3\mathrm{\pi }}{2x+6}dx=\frac{3\mathrm{\pi }}{2}-\mathrm{\pi }\left[\ln \left(8\right)\right]$.

Step-by-step solution

Step 1. Given information.

A definite integral is given as${\int }_1^{e-3}\frac{3\mathrm{\pi }}{2x+6}dx$.

Step 2. Using the Fundamental Theorem of Calculus.

We get

$$\begin{gathered} {\int }_1^{e-3}\frac{3\mathrm{\pi }}{2x+6}dx \\ =\frac{3\mathrm{\pi }}{2}{\int }_1^{e-3}\frac{1}{x+3}dx \\ =\frac{3\mathrm{\pi }}{2}{\left[\ln \right|x+3\left|\right]}_1^{e-3} \\ =\frac{3\mathrm{\pi }}{2}[\ln \left(e-3+3\right)-\ln \left(1+4\right)] \\ =\frac{3\mathrm{\pi }}{2}[\ln \left(e\right)-\ln \left(4\right)] \\ =\frac{3\mathrm{\pi }}{2}\ln \left(e\right)-\frac{3\mathrm{\pi }}{2}\ln \left(4\right) \\ =\frac{3\mathrm{\pi }}{2}-\frac{3\mathrm{\pi }}{2}\left(\frac{2}{3}\right)\ln \left(8\right) \\ =\frac{3\mathrm{\pi }}{2}-\mathrm{\pi ln}\left(8\right) \end{gathered}$$

The exact value of the given definite integral is$\frac{3\mathrm{\pi }}{2}-\pi \left[\ln \left(8\right)\right]$.

Step 3. The graph to verify the answer is 

The solution is area under graph which is

$$\begin{gathered} a\approx -1.820369 \\ \approx \frac{3\mathrm{\pi }}{2}-\mathrm{\pi }\left[\ln \left(8\right)\right] \end{gathered}$$