Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

8. Definite Integrals

Average Value of a Function

Calculus

8. Definite Integrals

Average Value of a Function

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2:45 minutes

Problem 5.4.39

Textbook Question

Mean Value Theorem for Integrals Find or approximate all points at which the given function equals its average value on the given interval.

ƒ(𝓍) = 8 ― 2𝓍 on [0, 4]

Verified step by step guidance

Step 1: Recall the formula for the average value of a function on an interval [a, b]. The average value is given by:

\frac{1}{(b - a)} \int a \to b ƒ (𝓍) d 𝓍

. Here, a = 0 and b = 4.

Step 2: Compute the definite integral of ƒ(𝓍) = 8 - 2𝓍 over the interval [0, 4]. Set up the integral:

\int 0 \to 4 (8 - 2 𝓍) d 𝓍

. Evaluate this integral step by step.

Step 3: Divide the result of the integral by the length of the interval (b - a = 4 - 0 = 4) to find the average value of the function on [0, 4]. This gives the average value of ƒ(𝓍).

Step 4: Set ƒ(𝓍) equal to the average value found in Step 3. Solve the equation

8 - 2 𝓍 = average_value

for 𝓍.

Step 5: Verify that the solution(s) for 𝓍 lie within the interval [0, 4]. These are the points at which the function equals its average value on the given interval.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function is continuous on a closed interval [a, b], then there exists at least one point c in (a, b) such that the function's value at c equals the average value of the function over that interval. This average value is calculated as (1/(b-a)) * ∫[a to b] f(x) dx.

Average Value of a Function

The average value of a function f(x) over the interval [a, b] is defined as (1/(b-a)) * ∫[a to b] f(x) dx. This concept allows us to determine a single representative value of the function across the interval, which can then be compared to the function's actual values at specific points within that interval.

Finding Points of Intersection

To find points where a function equals its average value, we set the function f(x) equal to the average value calculated from the previous concepts. This involves solving the equation f(x) = average value, which may require algebraic manipulation or numerical methods to identify the specific x-values where this equality holds.

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