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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 5.3.96a

Working with area functions Consider the function ƒ and the points a, b, and c.
(a) Find the area function A (𝓍) = ∫ₐˣ ƒ(t) dt using the Fundamental Theorem.
ƒ(𝓍) = ― 12𝓍 (𝓍―1) (𝓍― 2) ; a = 0 , b = 1 , c = 2

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1
Step 1: Recall the Fundamental Theorem of Calculus, which states that if A(𝓍) = ∫ₐˣ ƒ(t) dt, then A'(𝓍) = ƒ(𝓍). This means the area function A(𝓍) is the antiderivative of ƒ(t) evaluated from a to 𝓍.
Step 2: Write down the given function ƒ(𝓍) = -12𝓍(𝓍 - 1)(𝓍 - 2). To find A(𝓍), we need to integrate ƒ(t) with respect to t over the interval [a, 𝓍], where a = 0.
Step 3: Substitute ƒ(t) = -12t(t - 1)(t - 2) into the integral A(𝓍) = ∫ₐˣ ƒ(t) dt. This becomes A(𝓍) = ∫₀ˣ -12t(t - 1)(t - 2) dt.
Step 4: Expand the polynomial -12t(t - 1)(t - 2) to simplify the integrand. Multiply the terms step by step to get a single polynomial expression in terms of t.
Step 5: Integrate the resulting polynomial term by term with respect to t. Use the power rule for integration, ∫tⁿ dt = (tⁿ⁺¹)/(n+1), and evaluate the definite integral from 0 to 𝓍 to find A(𝓍).

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주요 개념

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Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration. It states that if a function is continuous on an interval, then the integral of its derivative over that interval can be computed using the values of the function at the endpoints. This theorem is essential for finding area functions, as it allows us to express the area under a curve as an integral.
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가이드 코스
06:11
Fundamental Theorem of Calculus Part 1

Definite Integral

A definite integral represents the signed area under a curve defined by a function over a specific interval [a, b]. It is calculated as the limit of Riemann sums and provides a numerical value that corresponds to the total accumulation of the function's values between the two bounds. In the context of area functions, it is used to compute the area from a starting point 'a' to a variable endpoint 'x'.
추천 영상:
가이드 코스
05:43
Definition of the Definite Integral

Area Function

An area function A(x) is defined as the integral of a function f(t) from a constant lower limit 'a' to a variable upper limit 'x'. It quantifies the area under the curve of f(t) from 'a' to 'x'. This function is crucial for understanding how the area changes as 'x' varies, and it is often used in applications involving accumulation and total quantities.
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05:06
Finding Area When Bounds Are Not Given
관련 실천
교과서 질문

Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.



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교과서 질문

{Use of Tech} Approximating definite integrals with a calculator Consider the following definite integrals.

(a) Write the left and right Riemann sums in sigma notation for an arbitrary value of n.


∫₀¹ cos ⁻¹ 𝓍 d𝓍

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교과서 질문

Properties of integrals Consider two functions ƒ and g on [1,6] such that ∫₁⁶ƒ(𝓍) d𝓍 = 10 and ∫₁⁶g(𝓍) d𝓍 = 5, ∫₄⁶ƒ(𝓍) d𝓍 = 5 , and ∫₁⁴g(𝓍) d𝓍 = 2. Evaluate the following integrals.


(a) ∫₁⁴ 3f(𝓍) d𝓍

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교과서 질문

The velocity in ft/s of an object moving along a line is given by v = ƒ(t) on the interval 0 ≤ t ≤ 8 (see figure), where t is measured in seconds.

a) Divide the interval [0,8] into n = 2 subintervals, [0,4] and [4,8]. On each subinterval, assume the object moves at a constant velocity equal to the value of v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0,8] (see part (a) of the figure)                                                                                                             


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교과서 질문

Area functions The graph of ƒ is shown in the figure. Let A(x) = ∫₋₂ˣ ƒ(t) dt and F(x) = ∫₄ˣ ƒ(t) dt be two area functions for ƒ. Evaluate the following area functions.

(a) A (―2)

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교과서 질문

Planetary orbits The planets orbit the Sun in elliptical orbits with the Sun at one focus (see Section 12.4 for more on ellipses). The equation of an ellipse whose dimensions are 2a in the 𝓍-direction and 2b in the y-direction is (𝓍²/a²) + (y² /b²) = 1.

(a) Let d² denote the square of the distance from a planet to the center of the ellipse at (0, 0). Integrate over the interval [ ―a, a] to show that the average value of d² is (a² + 2b²) /3 .

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