Gateway Arch The Gateway Arch in St. Louis is 630 ft high and has a 630-ft base. Its shape can be modeled by the parabola y = 630 (1― (𝓍/315)²) . Find the average height of the arch above the ground.
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 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
- 10. Physics Applications of Integrals 3h 16m
- 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
- 16. Parametric Equations & Polar Coordinates7h 58m
8. Definite Integrals
Average Value of a Function
Problem 5.4.3b
Textbook Question
Suppose ƒ is an even function and ∫⁸₋₈ ƒ(𝓍) d𝓍 = 18
(b) Evaluate ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍 .

1
Step 1: Recall the definition of an even function. An even function satisfies ƒ(𝓍) = ƒ(-𝓍) for all 𝓍 in its domain. This symmetry will be important for solving the integral.
Step 2: Consider the integral ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍. Notice that the integrand 𝓍ƒ(𝓍) involves the product of 𝓍 and ƒ(𝓍).
Step 3: Analyze the symmetry of the integrand. Since ƒ(𝓍) is even, ƒ(𝓍) = ƒ(-𝓍). However, 𝓍 is an odd function because 𝓍 = -𝓍 when reflected about the origin. The product of an even function and an odd function is an odd function.
Step 4: Recall a key property of definite integrals: the integral of an odd function over a symmetric interval [−a, a] is always 0. This is because the positive and negative contributions cancel each other out.
Step 5: Conclude that ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍 = 0 based on the symmetry of the integrand and the properties of odd functions over symmetric intervals.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Even Functions
An even function is defined as a function f(x) that satisfies the condition f(-x) = f(x) for all x in its domain. This symmetry about the y-axis implies that the area under the curve from -a to 0 is equal to the area from 0 to a. This property is crucial for evaluating integrals involving even functions, as it simplifies calculations and allows for certain symmetries to be exploited.
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Properties of Definite Integrals
Definite integrals have several important properties, one of which is that the integral of an odd function over a symmetric interval around zero is zero. This is because the areas above and below the x-axis cancel each other out. Understanding these properties helps in evaluating integrals more efficiently, especially when dealing with functions that exhibit symmetry.
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Integration by Parts
Integration by parts is a technique used to integrate products of functions and is based on the product rule for differentiation. It is expressed as ∫u dv = uv - ∫v du, where u and v are differentiable functions. This method is particularly useful when integrating products of polynomials and other functions, allowing for the transformation of complex integrals into simpler forms.
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