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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 5.5.98

Areas of regions Find the area of the following regions.                                                                                                                   
                                                                                                                                                                 The region bounded by the graph of ƒ(𝓍) = (𝓍―4)⁴ and the 𝓍-axis between and 𝓍 = 2 and 𝓍= 6

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Identify the function and the interval: The function is given as \(f(x) = (x - 4)^4\), and the region is bounded between \(x = 2\) and \(x = 6\) along the \(x\)-axis.
Determine the nature of the function on the interval: Since \((x - 4)^4\) is always non-negative for all real \(x\), the graph lies above or on the \(x\)-axis between \(x=2\) and \(x=6\).
Set up the definite integral to find the area: The area under the curve and above the \(x\)-axis from \(x=2\) to \(x=6\) is given by the integral \(\int_{2}^{6} (x - 4)^4 \, dx\).
Apply the power rule for integration: Recall that \(\int (x - a)^n \, dx = \frac{(x - a)^{n+1}}{n+1} + C\). Use this to write the antiderivative of \((x - 4)^4\).
Evaluate the definite integral: Substitute the upper limit \(x=6\) and the lower limit \(x=2\) into the antiderivative expression and find the difference to get the area.

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주요 개념

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

Definite Integral as Area Under a Curve

The definite integral of a function between two points gives the net area between the graph and the x-axis over that interval. For functions above the x-axis, this corresponds to the actual area. Calculating this integral helps find the exact area of the region bounded by the curve and the axis.
추천 영상:
가이드 코스
05:43
Definition of the Definite Integral

Properties of Even-Powered Polynomial Functions

Functions like ƒ(x) = (x - 4)^4 are always non-negative because even powers eliminate negative values. This ensures the graph lies on or above the x-axis, simplifying area calculations since the integral directly represents the area without needing to consider negative parts.
추천 영상:
가이드 코스
06:21
Properties of Functions

Setting Integration Limits Based on the Region

The limits of integration correspond to the x-values that bound the region of interest. Here, the area is between x = 2 and x = 6, so these values are used as the lower and upper limits in the definite integral to calculate the area under the curve within this interval.
추천 영상:
가이드 코스
07:06
Integration by Parts for Definite Integrals Example 8
관련 실천
교과서 질문

Multiple substitutions If necessary, use two or more substitutions to find the following integrals.                                                                                    

                                                                                                                                                                    

  ∫ 𝓍 sin⁴ 𝓍² cos 𝓍² d𝓍 (Hint: Begin with u = 𝓍², and then use v = sin u .)

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교과서 질문

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus


∫¹₁/₂ (t⁻³ ― 8) dt

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교과서 질문

Derivatives of integrals Simplify the following expressions.


d/dy ∫¹⁰ᵧ³ √(𝓍⁶ + 1) d𝓍

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교과서 질문

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.                                                                                                                         

                                                                                                                                                                              

 ∫π/₁₆^π/⁸ 8 csc² 4𝓍 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) = 1/(𝓍² + 1) on [―1, 1]

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교과서 질문

Integrals with sin² 𝓍 and cos² 𝓍 Evaluate the following integrals.                                                                                                             

                                                                                                                                                                    

 ∫₀^π/⁴ cos² 8θ dθ

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