Q. 60 For each function f and interval... [FREE SOLUTION]

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Chapter 4: Q. 60 (page 386)

For each function f and interval[a, b] in Exercises 56鈥�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.
$f (x) = {(x + 2)}^{2} 鈭� 5, [鈭� 5, 0]$

Short Answer

Expert verified

The exact average value of f is $- 2.6667,$ and it is verified from the graph of f.

The graph is

Step by step solution

Step 1. Given Information.

The given function and interval is $f (x) = {(x + 2)}^{2} 鈭� 5, [鈭� 5, 0] .$

Step 2. Finding the exact average value.

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

Thus,

localid="1649764183134" $\frac{1}{b - a} {鈭�}_{a}^{b} f (x) d x = \frac{1}{0 - (- 5)} {鈭�}_{- 5}^{0} {(x + 2)}^{2} - 5 d x = \frac{1}{5} {鈭�}_{- 5}^{0} (x^{2} + 4 x + 4 - 5) d x = \frac{1}{5} {鈭�}_{- 5}^{0} (x^{2} + 4 x - 1) d x = \frac{1}{5} {[\frac{x^{3}}{3} + \frac{4 x^{2}}{2} - x]}_{- 5}^{0} = \frac{1}{5} [\frac{0}{3} + \frac{4 (0)}{2} - 0 - (\frac{{(- 5)}^{3}}{3} + \frac{4 {(- 5)}^{2}}{2} - (- 5))] = \frac{1}{5} [0 - (- \frac{125}{3} + 50 + 5)] = \frac{1}{5} [\frac{125}{3} - 55] = \frac{1}{5} [\frac{125 - 165}{3}] = \frac{1}{5} (- \frac{40}{3}) = - \frac{8}{3} = - 2.6667$

Step 3. Verification.

By using the graphing utility, the graph of f is

From the graph, we can depict that the average value is $- 2.6667 .$ Thus, the answer is right.

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91影视

For each function f and interval[a, b] in Exercises 56鈥�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.
$f (x) = {(x + 2)}^{2} 鈭� 5, [鈭� 5, 0]$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Finding the exact average value.

Step 3. Verification.

One App. One Place for Learning.

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91影视

For each function f and interval[a, b] in Exercises 56鈥�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)=(x+2)2鈭�5,[鈭�5,0]

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Finding the exact average value.

Step 3. Verification.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Pure Maths

Statistics

Geometry

Mechanics Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

For each function f and interval[a, b] in Exercises 56鈥�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.
$f (x) = {(x + 2)}^{2} 鈭� 5, [鈭� 5, 0]$