Q. 67 - Calculus

Question

For each function f and interval [a, b] in Exercises 56–67, use definite integrals and the Fundamental Theorem of Calculus to find the exact average value of f from x = a to x = b. Then use a graph of f to verify that your answer is reasonable.  

$f\left(x\right)=\sin x+x \cos x, [a, b] = [-\pi , \pi ]$

Step-by-Step Solution

Answer

The exact average value of f  is$0,$and it is verified from the graph of f.

The graph is

Step 1. Given Information.

The given function and interval is $f\left(x\right)=\sin x+x \cos x, [a, b] = [-\pi , \pi ].$

Step 2. Finding the exact average value.

To find the exact average value of f  from $x=a to x=b,$ we will use the formula: $\frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx.$

Thus,

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

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that the average value is $0.$ Thus, the answer is right.