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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.2.87
Capitolo 5, Problema 5.2.87

Area by geometry Use geometry to evaluate the following integrals.


∫⁴₋₆ √(24 ― 2𝓍 ― 𝓍²) d𝓍

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First, recognize that the integral involves the expression under the square root: \(24 - 2x - x^2\). To use geometry, rewrite this quadratic expression in a more recognizable form by completing the square.
Rewrite \(24 - 2x - x^2\) as \(-(x^2 + 2x - 24)\). Then complete the square for the expression inside the parentheses: \(x^2 + 2x - 24 = (x^2 + 2x + 1) - 1 - 24 = (x + 1)^2 - 25\).
Substitute back to get \(24 - 2x - x^2 = -( (x + 1)^2 - 25 ) = 25 - (x + 1)^2\). So the integral becomes \(\int_{-6}^4 \sqrt{25 - (x + 1)^2} \, dx\).
Interpret the integral geometrically: \(\sqrt{25 - (x + 1)^2}\) represents the upper half of a circle centered at \(x = -1\) with radius \(5\). The integral from \(x = -6\) to \(x = 4\) corresponds to the area under this semicircle between these limits.
Calculate the area of the circular segment corresponding to the interval \([-6, 4]\) by finding the area of the semicircle (or relevant sector) and subtracting any areas outside the integration bounds, using geometric formulas for circle segments.

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Interpreting Integrals as Areas

Definite integrals can represent the area under a curve between two points on the x-axis. When the integrand corresponds to a geometric shape, the integral's value equals the area of that shape, allowing evaluation without direct integration.
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Finding Area When Bounds Are Not Given

Completing the Square

Completing the square rewrites quadratic expressions into a form (x - h)² + k, revealing geometric shapes like circles or parabolas. This technique helps identify the curve's shape and its parameters, facilitating area calculation using geometry.
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Completing the Square to Rewrite the Integrand

Area of a Circle Segment

When the integrand describes a semicircle or circle segment, the area can be found using formulas for circle areas or segments. Recognizing the radius and center from the equation allows direct computation of the integral as a geometric area.
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Finding Area When Bounds Are Not Given
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