Q. 66 - 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.  

$f\left(x\right)=x^2\sin (x^3+1), [a, b] = [-1, 2]$

Step-by-Step Solution

Answer

The exact average value of f is $0.2123$ and it is verified from the graph of f.

Step 1. Given Information.

The given function and interval is $f\left(x\right)=x^2\sin (x^3+1), [a, b] = [-1, 2].$

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,

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{2-(-1)}{\int }_{-1}^2{x}^2\sin \left(x^3+1\right) dx \\ =\frac{1}{3}\left(-\frac{1}{3}\right){\left[\cos (x^3+1)\right]}_{-1}^2 \\ =-\frac{1}{9}\left[\cos \left(9\right)-\cos \left(0\right)\right] \\ =-\frac{1}{9}\left[\cos \left(9\right)-1\right] \\ =-\frac{1}{9}\cos \left(9\right)+\frac{1}{9} \\ \approx 0.2123 \end{gathered}$$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

 

From the graph, we can depict that the average value is $0.2123.$ Thus, the answer is right.