Q. 46 - Calculus

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.)

${\int }_{-3}^{3}2\cos \left(\mathrm{\pi x}\right)dx$

Step-by-Step Solution

Answer

${\int }_{-3}^{3}2\cos \left(\mathrm{\pi x}\right)dx=0$.

Step 1. Given information.

A definite integral is given as ${\int }_{-3}^{3}2\cos \left(\mathrm{\pi x}\right)dx$.

Step 2. Using the Fundamental Theorem of Calculus.

We get

$$\begin{gathered} {\int }_{-3}^{3}2\cos \left(\mathrm{\pi x}\right)dx \\ =2{\int }_{-3}^3\cos \left(\mathrm{\pi x}\right)dx \\ =2\left(\frac{1}{\mathrm{\pi }}\right){\left[\sin \right(\mathrm{\pi x}\left)\right]}_{-3}^3 \\ =\frac{2}{\mathrm{\pi }}[\sin 3\mathrm{\pi }-\sin (-3\left)\mathrm{\pi }\right] \\ =\frac{2}{\mathrm{\pi }}(\sin 3\mathrm{\pi }+\sin 3\mathrm{\pi }) \\ =\frac{2}{\mathrm{\pi }}\sin 6\mathrm{\pi } \\ =\frac{2}{\mathrm{\pi }}\left(0\right) \\ =0 \end{gathered}$$

The exact value of the given definite integral is 0.

Step 3. The graph to verify the answer is

The solution is area under graph which is

$a=0$.