Question

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

${\int }_{-\pi }^{\pi } (1+\sin x)dx$

Short answer

Ans: The exact value of${\int }_{-\pi }^{\pi } (1+\sin x)dx=2\mathrm{\pi }$

Step-by-step solution

Step 1. Given information.

given,

${\int }_{-\pi }^{\pi } (1+\sin x)dx$

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

The exact value is calculated as shown below,

$\begin{array}{r}{\int }_{-\pi }^{\pi } (1+\sin x)dx\\ ={\int }_{-\pi }^{\pi } \left(1\right)dx+{\int }_{-\pi }^{\pi } (\sin x)dx\\ =[x{]}_{-\pi }^{\pi }+[-\cos x{]}_{-\pi }^{\pi }\\ =\pi +\pi -\cos \pi +\cos (-\pi )\\ =2\pi +0\\ =2\pi \end{array}$

Therefore, the exact value is, $2\mathrm{\pi }.$

Step 3. Check the answer using a graph.

The required graph is,