Textbook Question
Matching functions with area functions Match the functions Ζ, whose graphs are given in aβ d, with the area functions A (π) = β«βΛ£ Ζ(t) dt, whose graphs are given in AβD.
49
views
Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.3.96a
Working with area functions Consider the function Ζ and the points a, b, and c.
(a) Find the area function A (π) = β«βΛ£ Ζ(t) dt using the Fundamental Theorem.
Ζ(π) = β 12π (πβ1) (πβ 2) ; a = 0 , b = 1 , c = 2
Verified step by step guidance1
Step 1: Recall the Fundamental Theorem of Calculus, which states that if A(π) = β«βΛ£ Ζ(t) dt, then A'(π) = Ζ(π). This means the area function A(π) is the antiderivative of Ζ(t) evaluated from a to π.
Step 2: Write down the given function Ζ(π) = -12π(π - 1)(π - 2). To find A(π), we need to integrate Ζ(t) with respect to t over the interval [a, π], where a = 0.
Step 3: Substitute Ζ(t) = -12t(t - 1)(t - 2) into the integral A(π) = β«βΛ£ Ζ(t) dt. This becomes A(π) = β«βΛ£ -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, β«tβΏ dt = (tβΏβΊΒΉ)/(n+1), and evaluate the definite integral from 0 to π to find A(π).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
Recommended video:
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'.
Recommended video:
05:43Definition 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.
Recommended video:
05:06Finding Area When Bounds Are Not Given
Related Practice
Textbook Question
{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.
β«βΒΉ cos β»ΒΉ π dπ
57
views
Textbook Question
Properties of integrals Consider two functions Ζ and g on [1,6] such that β«ββΆΖ(π) dπ = 10 and β«ββΆg(π) dπ = 5, β«ββΆΖ(π) dπ = 5 , and β«ββ΄g(π) dπ = 2. Evaluate the following integrals.
(a) β«ββ΄ 3f(π) dπ
61
views
Textbook Question
The velocity in ft/s of an object moving along a line is given by v = Ζ(t) on the interval 0 β€ t β€ 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)
89
views
Textbook Question
Area functions The graph of Ζ is shown in the figure. Let A(x) = β«ββΛ£ Ζ(t) dt and F(x) = β«βΛ£ Ζ(t) dt be two area functions for Ζ. Evaluate the following area functions.
(a) A (β2)
56
views
Textbook Question
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 π-direction and 2b in the y-direction is (πΒ²/aΒ²) + (yΒ² /bΒ²) = 1.
(a) Let dΒ² denote the square of the distance from a planet to the center of the ellipse at (0, 0). Integrate over the interval [ βa, a] to show that the average value of dΒ² is (aΒ² + 2bΒ²) /3 .
137
views