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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.2.43
Chapter 5, Problem 5.2.43

Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result.                                                                                                                                      
                                                                                                                                                                                       
 ∫₀⁴ √(16― 𝓍² ) d𝓍

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Recognize that the integral \(\int_0^4 \sqrt{16 - x^2} \, dx\) represents the area under the curve \(y = \sqrt{16 - x^2}\) from \(x=0\) to \(x=4\).
Note that the equation \(y = \sqrt{16 - x^2}\) describes the upper half of a circle centered at the origin with radius 4, since \(x^2 + y^2 = 16\).
Sketch the circle \(x^2 + y^2 = 16\) and shade the region from \(x=0\) to \(x=4\) under the upper semicircle to visualize the area represented by the integral.
Calculate the area of the quarter circle (since \(x\) goes from 0 to 4 and \(y\) is positive) using the formula for the area of a circle sector: \(\text{Area} = \frac{1}{4} \pi r^2\) where \(r=4\).
Conclude that the value of the definite integral equals the area of this quarter circle, which is \(\frac{1}{4} \pi \times 4^2\).

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Key Concepts

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

Definite Integral as Area Under a Curve

A definite integral represents the net area between the graph of a function and the x-axis over a specified interval. When the function is non-negative, this corresponds to the geometric area under the curve. Understanding this allows one to interpret integrals as areas, which can be calculated using geometric formulas instead of limits or sums.
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Geometry of a Circle and Its Equation

The integrand √(16 - x²) describes the upper half of a circle centered at the origin with radius 4, since x² + y² = 16 is the equation of a circle. Recognizing this helps in visualizing the region under the curve as a semicircle, enabling the use of geometric area formulas to evaluate the integral.
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Using Geometric Formulas to Evaluate Integrals

Instead of computing the integral via Riemann sums or antiderivatives, one can use known geometric area formulas, such as the area of a semicircle (½πr²), to find the value of the integral. This approach simplifies the problem by connecting calculus with classical geometry.
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