Q. 61 - 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.  

$f\left(x\right)=x^2-2x-1, [0, 3]$

Step-by-Step Solution

Answer

The exact average value of f is $-1$ and it is verified from the graph of f. 

Step 1. Given Information.

The given function and interval is $f\left(x\right)=x^2-2x-1, [0, 3].$

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,

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{3-0}{\int }_0^3{x}^2-2x-1 dx \\ =\frac{1}{3}{\left[\frac{x^3}{3}-\frac{2x^2}{2}-x\right]}_0^3 \\ =\frac{1}{3}\left[\frac{3^3}{3}-\frac{2{\left(3\right)}^2}{2}-3-\left(\frac{0}{3}-\frac{2\left(0\right)}{2}-0\right)\right] \\ =\frac{1}{3}\left[9-9-3-0\right] \\ =-1 \end{gathered}$$

Step 3. Verification.

The graph of f  is drawn in the given interval.

From the graph, we can observe that the area between the curve and the x-axis is negative. As the larger bounded portion is below the x -axis. The average value is $-1$.