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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3:05 minutes

Problem 5.4.4a

Textbook Question

Suppose ƒ is an odd function, ∫₀⁴ ƒ(𝓍) d𝓍 = 3 , and ∫₀⁸ ƒ(𝓍) d𝓍 = 9 .

(a) Evaluate ∫₋₈⁴ ƒ(𝓍) d𝓍 .

Verified step by step guidance

Step 1: Recall the property of odd functions. An odd function satisfies ƒ(−𝓍) = −ƒ(𝓍). This property implies that the integral of an odd function over a symmetric interval [−𝓎, 𝓎] is zero, i.e., ∫₋𝓎⁺𝓎 ƒ(𝓍) d𝓍 = 0.

Step 2: Break the integral ∫₋₈⁴ ƒ(𝓍) d𝓍 into two parts using the additive property of integrals: ∫₋₈⁴ ƒ(𝓍) d𝓍 = ∫₋₈⁰ ƒ(𝓍) d𝓍 + ∫₀⁴ ƒ(𝓍) d𝓍.

Step 3: Evaluate ∫₋₈⁰ ƒ(𝓍) d𝓍 using the property of odd functions. Since the interval [−8, 0] is symmetric about zero, ∫₋₈⁰ ƒ(𝓍) d𝓍 = −∫₀⁸ ƒ(𝓍) d𝓍. Substitute the given value ∫₀⁸ ƒ(𝓍) d𝓍 = 9 to find ∫₋₈⁰ ƒ(𝓍) d𝓍 = −9.

Step 4: Substitute the given value of ∫₀⁴ ƒ(𝓍) d𝓍 = 3 into the equation from Step 2: ∫₋₈⁴ ƒ(𝓍) d𝓍 = −9 + 3.

Step 5: Combine the results from Step 4 to express the final integral value. The result is ∫₋₈⁴ ƒ(𝓍) d𝓍 = −6.

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Key Concepts

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

Odd Functions

An odd function is defined by the property that ƒ(-x) = -ƒ(x) for all x in its domain. This symmetry about the origin implies that the area under the curve from -a to 0 is the negative of the area from 0 to a. Understanding this property is crucial for evaluating integrals involving odd functions, especially when considering limits that span both positive and negative values.

Definite Integrals

A definite integral, denoted as ∫ₐᵇ ƒ(x) dx, represents the signed area under the curve of the function ƒ(x) from x = a to x = b. The value of a definite integral can be interpreted as the accumulation of the function's values over the specified interval. In this problem, the given integrals provide specific areas that can be used to find other related integrals through properties of symmetry and linearity.

Properties of Integrals

The properties of integrals, such as the linearity of integration and the relationship between integrals over symmetric intervals, are essential for solving problems involving definite integrals. For instance, the integral from -a to a of an odd function is zero, and the integral from a to b can be expressed in terms of integrals over other intervals. These properties allow for simplifications and transformations that facilitate the evaluation of complex integrals.

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