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The figure shows a parabolic segment, that is, a portion of a parabola cut off by a chord AB. It also shows a point C on the parabola with the property that the tangent line at C is parallel to the chord AB. Archimedes proved that the area of the parabolic segment is times the area of the inscribed triangle ABC. Verify Archimedes' result for the parabola y = 4 - x² and the line y = x + 2. 3 The parabola y = 4 - x² and the line y = x + 2 intersect when x = -2 or 1. So the point A is (x, y) = and the slope of the parabola y = 4 - x² at x-coordinate x is The line through A and C has slope y = - 1 +> B + 2) dx = 2 i The area A₁ of the parabolic segment is the area under the parabola from x = -2 to x = 1, minus the area under the line y = x + 2 from -2 to 1. Thus, we get the following. A₁ = - 1²(4-x²) dx - < - 1/²₁x + ² The area A₂ of the inscribed triangle is the area under the line segment AC plus the area under the line segment CB minus the area under the line segment AB. and equation y - 0 =. = 15/2/(x+ (x + 2), or y = and B is (x, y) = | Thus, we get the following. ^₂ = √ ( ² x + 5) ∞x + ( - 2 x + ²) ²x − L, ₁x + 2) ²x - [ (x dx £² dx - dx = . These slopes are equal when x = The slope of the line y = x + 2 is 5 x + 5. The line through C and B has slope , so the point C is (x, y) = and equation y - 3 = - -=1/(x. (x - 1), or

The figure shows a parabolic segment, that is, a portion of a parabola cut off by a chord AB. It also shows a point C on the parabola with the property that the tangent line at C is parallel to the chord AB. Archimedes proved that the area of the parabolic segment is times the area of the inscribed triangle ABC. Verify Archimedes' result for the parabola y = 4 - x² and the line y = x + 2. 3 The parabola y = 4 - x² and the line y = x + 2 intersect when x = -2 or 1. So the point A is (x, y) = and the slope of the parabola y = 4 - x² at x-coordinate x is The line through A and C has slope y = - 1 +> B + 2) dx = 2 i The area A₁ of the parabolic segment is the area under the parabola from x = -2 to x = 1, minus the area under the line y = x + 2 from -2 to 1. Thus, we get the following. A₁ = - 1²(4-x²) dx - < - 1/²₁x + ² The area A₂ of the inscribed triangle is the area under the line segment AC plus the area under the line segment CB minus the area under the line segment AB. and equation y - 0 =. = 15/2/(x+ (x + 2), or y = and B is (x, y) = | Thus, we get the following. ^₂ = √ ( ² x + 5) ∞x + ( - 2 x + ²) ²x − L, ₁x + 2) ²x - [ (x dx £² dx - dx = . These slopes are equal when x = The slope of the line y = x + 2 is 5 x + 5. The line through C and B has slope , so the point C is (x, y) = and equation y - 3 = - -=1/(x. (x - 1), or

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The figure shows a parabolic segment, that is, a portion of a parabola cut off by a chord AB. It also shows a point C on the parabola with the property that the tangent line at C is parallel to the chord AB. Archimedes proved that the area of the parabolic segment is times the area of the inscribed triangle ABC. Verify Archimedes' result for the parabola y = 4 - x² and the line y = x + 2. 3 The parabola y = 4 - x² and the line y = x + 2 intersect when x = -2 or 1. So the point A is (x, y) = and the slope of the parabola y = 4 - x² at x-coordinate x is The line through A and C has slope y = - 1 +> B + 2) dx = 2 i The area A₁ of the parabolic segment is the area under the parabola from x = -2 to x = 1, minus the area under the line y = x + 2 from -2 to 1. Thus, we get the following. A₁ = - 1²(4-x²) dx - < - 1/²₁x + ² The area A₂ of the inscribed triangle is the area under the line segment AC plus the area under the line segment CB minus the area under the line segment AB. and equation y - 0 =. = 15/2/(x+ (x + 2), or y = and B is (x, y) = | Thus, we get the following. ^₂ = √ ( ² x + 5) ∞x + ( - 2 x + ²) ²x − L, ₁x + 2) ²x - [ (x dx £² dx - dx = . These slopes are equal when x = The slope of the line y = x + 2 is 5 x + 5. The line through C and B has slope , so the point C is (x, y) = and equation y - 3 = - -=1/(x. (x - 1), or
Transcribed Image Text:The figure shows a parabolic segment, that is, a portion of a parabola cut off by a chord AB. It also shows a point C on the parabola with the property that the tangent line at C is parallel to the chord AB. Archimedes proved that the area of the parabolic segment is times the area of the inscribed triangle ABC. Verify Archimedes' result for the parabola y = 4 - x² and the line y = x + 2. 3 The parabola y = 4 - x² and the line y = x + 2 intersect when x = -2 or 1. So the point A is (x, y) = and the slope of the parabola y = 4 - x² at x-coordinate x is The line through A and C has slope y = - 1 +> B + 2) dx = 2 i The area A₁ of the parabolic segment is the area under the parabola from x = -2 to x = 1, minus the area under the line y = x + 2 from -2 to 1. Thus, we get the following. A₁ = - 1²(4-x²) dx - < - 1/²₁x + ² The area A₂ of the inscribed triangle is the area under the line segment AC plus the area under the line segment CB minus the area under the line segment AB. and equation y - 0 =. = 15/2/(x+ (x + 2), or y = and B is (x, y) = | Thus, we get the following. ^₂ = √ ( ² x + 5) ∞x + ( - 2 x + ²) ²x − L, ₁x + 2) ²x - [ (x dx £² dx - dx = . These slopes are equal when x = The slope of the line y = x + 2 is 5 x + 5. The line through C and B has slope , so the point C is (x, y) = and equation y - 3 = - -=1/(x. (x - 1), or
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