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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
5장, 문제 5.2.29

Area versus net area Graph the following functions. Then use geometry (not Riemann sums) to find the area and the net area of the region described.
The region between the graph of y = 1 - |x| and the x-axis, for -2 ≤ x ≤ 2

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First, understand the function given: \(y = 1 - |x|\). This is a V-shaped graph with its vertex at \((0,1)\) and it intersects the x-axis where \(1 - |x| = 0\).
Find the points where the graph intersects the x-axis by solving \(1 - |x| = 0\). This gives \(|x| = 1\), so the points are \(x = -1\) and \(x = 1\).
Sketch the graph between \(x = -2\) and \(x = 2\). Note that for \(|x| > 1\), the function \(y = 1 - |x|\) is negative, so the graph lies below the x-axis on the intervals \([-2, -1]\) and \([1, 2]\), and above the x-axis on \([-1, 1]\).
To find the net area, calculate the definite integral of \(y = 1 - |x|\) from \(-2\) to \(2\). Since the function changes sign, the net area is the sum of the positive area above the x-axis minus the area below the x-axis.
To find the total (or geometric) area, calculate the area of the two triangles formed above the x-axis on \([-1, 1]\) and the two triangles below the x-axis on \([-2, -1]\) and \([1, 2]\), then add their absolute values together.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Net Area

Net area refers to the integral of a function over an interval, accounting for areas above the x-axis as positive and below as negative. It represents the algebraic sum of areas, which can result in cancellation when parts lie below the x-axis.
추천 영상:
05:06
Finding Area When Bounds Are Not Given

Area Between a Curve and the x-axis

The area between a curve and the x-axis is the total size of the region bounded by the graph and the axis, always taken as positive. When the function dips below the x-axis, the area is found by integrating the absolute value or by geometric methods considering separate regions.
추천 영상:
05:23
Finding Area Between Curves on a Given Interval

Using Geometry to Find Areas

Instead of integration, geometric shapes like triangles and trapezoids can be used to find areas under curves when the graph forms simple shapes. This approach involves identifying shapes formed by the function and the x-axis and calculating their areas using known formulas.
추천 영상:
05:06
Finding Area When Bounds Are Not Given
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교과서 질문

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 ∫ [(√𝓍 + 1)⁴ / 2√𝓍 d𝓍

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교과서 질문

Average values Find the average value of the following functions on the given interval. Draw a graph of the function and indicate the average value.

ƒ(𝓍) = 𝓍ⁿ on [0,1] , for any positive integer n

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교과서 질문

Average values Find the average value of the following functions on the given interval. Draw a graph of the function and indicate the average value.

ƒ(𝓍) = 𝓍³ on [―1, 1]

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교과서 질문

Definite integrals from graphs The figure shows the areas of regions bounded by the graph of ƒ and the 𝓍-axis. Evaluate the following integrals.


∫ₐ⁰ ƒ(𝓍) d𝓍

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교과서 질문

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 ∫ 𝓍eˣ² d𝓍

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교과서 질문

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 ∫ 𝓍³ (𝓍⁴ + 16)⁶ d𝓍

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