Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.R.102e

Chapter 5, Problem 5.R.102e

Function defined by an integral Let H (𝓍) = ∫₀ˣ √(4 ― t²) dt, for ― 2 ≤ 𝓍 ≤ 2.
(e) Find the value of s such that H (𝓍) = sH(―𝓍)

Verified step by step guidance

Step 1: Understand the problem. The function H(𝓍) is defined as an integral from 0 to 𝓍 of √(4 - t²) dt. The goal is to find the value of s such that H(𝓍) = sH(-𝓍). This involves symmetry properties of the integral and the function.

Step 2: Analyze the integrand √(4 - t²). This function is even because √(4 - t²) = √(4 - (-t)²). This means the integrand is symmetric about the y-axis.

Step 3: Use the property of definite integrals for even functions. For an even function f(t), ∫₀ˣ f(t) dt is equal to ∫₋ˣ⁰ f(t) dt. This symmetry will help relate H(𝓍) and H(-𝓍).

Step 4: Express H(-𝓍) using the definition of the integral. H(-𝓍) = ∫₀⁻ˣ √(4 - t²) dt. By reversing the limits of integration, this becomes H(-𝓍) = -∫₋ˣ⁰ √(4 - t²) dt.

Step 5: Combine the symmetry property and the reversed integral to find the relationship between H(𝓍) and H(-𝓍). Use this relationship to determine the value of s such that H(𝓍) = sH(-𝓍).

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Key Concepts

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

Definite Integral

A definite integral represents the signed area under a curve defined by a function over a specific interval. In this case, H(𝓍) is defined as the integral of √(4 - t²) from 0 to 𝓍, which geometrically corresponds to the area under the curve of the function from the lower limit to the upper limit.

Symmetry in Functions

Symmetry in functions refers to the property where a function exhibits the same behavior when its input is negated. For the function H(𝓍), understanding its symmetry can help in finding the relationship between H(𝓍) and H(−𝓍), which is crucial for solving the equation H(𝓍) = sH(−𝓍).

Parameterization

Parameterization involves expressing a function in terms of a variable that can take on a range of values. In this context, the variable s acts as a parameter that scales the relationship between H(𝓍) and H(−𝓍). Finding the appropriate value of s requires analyzing how these two integrals relate to each other.

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