Textbook Question
Area functions from graphs The graph of ƒ is given in the figure. A(𝓍) = ∫₀ˣ ƒ(t) dt and evaluate A(2), A(5), A(8), and A(12).
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Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.2.87
Area by geometry Use geometry to evaluate the following integrals.
∫⁴₋₆ √(24 ― 2𝓍 ― 𝓍²) d𝓍
Verified step by step guidance1
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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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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05:06
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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05:22Completing 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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05:06Finding Area When Bounds Are Not Given
Related Practice
Textbook Question
Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus
∫₁⁴ (𝓍 ― 2)/√𝓍 d𝓍
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Textbook Question
Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.
∫ 𝓍 csc 𝓍² cot 𝓍² d𝓍
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Textbook Question
Integrals with sin² 𝓍 and cos² 𝓍 Evaluate the following integrals.
∫₋π^π cos² 𝓍 d𝓍
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Textbook Question
On which derivative rule is the Substitution Rule based?
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Textbook Question
{Use of Tech} Areas of regions Find the area of the region 𝑅 bounded by the graph of ƒ and the 𝓍-axis on the given interval. Graph ƒ and show the region 𝑅.
ƒ(𝓍) = 𝓍² (𝓍 ― 2) on [ ―1 , 3]
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