If ƒ is an even function, why is ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 = 2 ∫₀ᵃ ƒ(𝓍) d𝓍?

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Step 1: Recall the definition of an even function. A function ƒ(𝓍) is even if ƒ(𝓍) = ƒ(-𝓍) for all values of 𝓍 in its domain. This symmetry property will be key to understanding the integral relationship.

Step 2: Consider the integral ∫ᵃ₋ₐ ƒ(𝓍) d𝓍. This represents the area under the curve of ƒ(𝓍) from -a to a. Due to the symmetry of even functions, the area from -a to 0 is identical to the area from 0 to a.

Step 3: Break the integral ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 into two parts: ∫₋ₐ⁰ ƒ(𝓍) d𝓍 + ∫₀ᵃ ƒ(𝓍) d𝓍. Using the property of even functions, the integral over [-a, 0] is equal to the integral over [0, a].

Step 4: Replace ∫₋ₐ⁰ ƒ(𝓍) d𝓍 with ∫₀ᵃ ƒ(𝓍) d𝓍 because the areas are equal due to symmetry. This gives ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 = ∫₀ᵃ ƒ(𝓍) d𝓍 + ∫₀ᵃ ƒ(𝓍) d𝓍.

Step 5: Combine the two identical integrals to get ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 = 2 ∫₀ᵃ ƒ(𝓍) d𝓍. This shows that the integral over the symmetric interval [-a, a] is twice the integral over [0, a].

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주요 개념

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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 means that the function's values are the same for both positive and negative inputs. Understanding this property is crucial for evaluating integrals over symmetric intervals.

Definite Integrals

A definite integral, denoted as ∫ᵇₐ f(x) dx, represents the signed area under the curve of the function f(x) from x = a to x = b. When evaluating integrals of even functions over symmetric intervals, the area from -a to 0 is equal to the area from 0 to a, which is a key aspect in simplifying the integral.

Properties of Integrals

One important property of integrals is that for an even function f(x), the integral over a symmetric interval can be expressed as twice the integral from 0 to a. This is mathematically represented as ∫ᵃ₋ₐ f(x) dx = 2 ∫₀ᵃ f(x) dx, allowing for simplification in calculations and demonstrating the relationship between the areas under the curve.