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.102a

Chapter 5, Problem 5.R.102a

Function defined by an integral Let H (𝓍) = ∫₀ˣ √(4 ― t²) dt, for ― 2 ≤ 𝓍 ≤ 2.
(a) Evaluate H (0) .

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. For part (a), we are tasked with evaluating H(0), which means substituting 𝓍 = 0 into the integral.

Step 2: Substitute 𝓍 = 0 into the integral definition. This gives H(0) = ∫₀⁰ √(4 − t²) dt.

Step 3: Recognize that the integral's limits are both 0. When the upper and lower limits of an integral are the same, the integral evaluates to 0. This is because there is no interval over which to integrate.

Step 4: Conclude that H(0) = 0 based on the property of definite integrals where the limits are equal.

Step 5: Reflect on the result. This step demonstrates the importance of understanding integral properties and how they simplify calculations in specific cases.

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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. Evaluating a definite integral involves finding the antiderivative of the integrand and applying the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then the definite integral of f from a to b is F(b) - F(a). This theorem allows us to evaluate H(𝓍) by finding an antiderivative of the integrand and substituting the limits of integration.

Geometric Interpretation of Integrals

Integrals can be interpreted geometrically as the area under a curve. For the function √(4 - t²), which describes a semicircle, the integral from 0 to 𝓍 gives the area of the corresponding segment of the semicircle. Understanding this geometric interpretation aids in visualizing the result of the integral and its evaluation at specific points like H(0).

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Function defined by an integral Let H (𝓍) = ∫₀ˣ √(4 ― t²) dt, for ― 2 ≤ 𝓍 ≤ 2.(a) Evaluate H (0) .

Verified video answer for a similar problem:

Key Concepts

Definite Integral

Fundamental Theorem of Calculus

Geometric Interpretation of Integrals

Function defined by an integral Let H (𝓍) = ∫₀ˣ √(4 ― t²) dt, for ― 2 ≤ 𝓍 ≤ 2.
(a) Evaluate H (0) .