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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 5.2.65

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.
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∫ₐ⁰ ƒ(𝓍) d𝓍

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1
Identify the integral to evaluate: \(\int_{a}^{0} f(x) \, dx\). Notice that the limits of integration go from \(a\) to \(0\), which is in the reverse order of the usual left-to-right direction on the x-axis.
Recall the property of definite integrals that reversing the limits changes the sign: \(\int_{a}^{0} f(x) \, dx = -\int_{0}^{a} f(x) \, dx\).
Look at the graph and observe the area between \(x=0\) and \(x=a\). The shaded region above the x-axis has an area of 16, so \(\int_{0}^{a} f(x) \, dx = 16\) because the function is positive there.
Use the property from step 2 to write \(\int_{a}^{0} f(x) \, dx = -16\) since the integral from 0 to a is positive 16 but the limits are reversed.
Thus, the value of the integral \(\int_{a}^{0} f(x) \, dx\) corresponds to the negative of the area between \(0\) and \(a\) under the curve \(f(x)\).

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주요 개념

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

Definite Integral as Net Area

A definite integral over an interval represents the net area between the function's graph and the x-axis. Areas above the x-axis contribute positively, while areas below contribute negatively. This net area interpretation is essential for evaluating integrals from graphs.
추천 영상:
05:43
Definition of the Definite Integral

Interpreting Areas from Graphs

When given shaded areas on a graph, these represent the absolute values of integrals over subintervals. To find the integral over a larger interval, sum these areas with appropriate signs based on whether the function is above or below the x-axis.
추천 영상:
06:15
Graphing The Derivative

Properties of Definite Integrals and Limits

The integral from a to 0 can be evaluated by reversing limits: ∫ₐ⁰ f(x) dx = -∫₀ᵃ f(x) dx. Understanding how to manipulate integral limits and combine subinterval integrals is crucial for solving problems involving integrals over multiple segments.
추천 영상:
05:43
Definition of the Definite Integral
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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

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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교과서 질문

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. Sketch the graph of the integrand and shade the region whose net area you have found.                       

                                                                                                                                                                                       

 ∫₀⁵ (𝓍²―9) 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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