Q. 64

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.

$f (x) = \frac{1}{3 x + 1}, [a, b] = [2, 5]$

Step-by-Step Solution

Verified

Answer

The exact average value of f is $0.0919,$ and it is verified from the graph of f.

The graph is

1Step 1. Given Information.

The given function and interval is $f (x) = \frac{1}{3 x + 1}, [a, b] = [2, 5] .$

2Step 2. Finding the exact average value.

To find the exact average value of f from $x = a t o x = b,$ we will use the formula: $\frac{1}{b - a} \int_{a}^{b} f (x) d x .$

Thus,

$\frac{1}{b - a} \int_{a}^{b} f (x) d x = \frac{1}{5 - 2} \int_{2}^{5} \frac{1}{3 x + 1} d x = \frac{1}{3} {[\frac{1}{3} \ln |3 x + 1|]}_{2}^{5} = \frac{1}{9} {[\ln |3 x + 1|]}_{2}^{5} = \frac{1}{9} [\ln |3 (5) + 1| - \ln |3 (2) + 1|] = \frac{1}{9} [\ln (16) - \ln (7)] = 0.0919$

3Step 3. Verification.

By using the graphing utility, the graph of f is

From the graph, we can depict that the average value is $0.0919 .$ Thus, the answer is right.

Q. 63

Q. 65

Other exercises in this chapter

Q. 62

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

View solution

Q. 63

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 av

View solution

Q. 65

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

View solution

Q. 66

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

View solution