Textbook Question
Area functions for linear functions Consider the following functions Ζ and real numbers a (see figure).
(a) Find and graph the area function A (π) = β«βΛ£ Ζ(t) dt .
Ζ(t) = 2t + 5 , a = 0
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Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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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 5.3.97a
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) = cos π ; a = 0 , b = Ο/2 , c = Ο
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 derivative of the area function A(π) is equal to the original function Ζ(π).
Step 2: To find the area function A(π), integrate Ζ(t) = cos(t) with respect to t from the lower limit a = 0 to the upper limit π. The integral of cos(t) is sin(t).
Step 3: Apply the definite integral formula: A(π) = β«βΛ£ Ζ(t) dt = [sin(t)]βΛ£. Substitute the limits of integration into the antiderivative.
Step 4: Substitute the lower limit a = 0 and the upper limit π into the expression: A(π) = sin(π) - sin(0). Simplify the result using the fact that sin(0) = 0.
Step 5: The area function A(π) is now expressed as A(π) = sin(π). This function represents the accumulated area under Ζ(t) = cos(t) from t = 0 to t = π.
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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 connects differentiation and integration, stating that if a function is continuous on an interval, then the integral of its derivative over that interval gives the net change of the function. Specifically, it allows us to evaluate definite integrals using antiderivatives, which is essential for finding area functions.
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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 bounds, which is crucial for determining the area function A(x) in the given problem.
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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. This function represents the accumulated area under the curve of f(t) from a to x, and it is essential for understanding how the area changes as x varies, particularly in the context of the problem involving the cosine function.
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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
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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
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Textbook Question
Average value with a parameter Consider the function Ζ(π) = aπ (1βπ) on the interval [0, 1], where a is a positive real number.
(a) Find the average value of Ζ as a function of a .
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Textbook Question
Mass from density A thin 10-cm rod is made of an alloy whose density varies along its length according to the function shown in the figure. Assume density is measured in units of g/cm. In Chapter 6, we show that the mass of the rod is the area under the density curve.
(a) Find the mass of the left half of the rod (0 β€ x β€ 5) .
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Textbook Question
Approximating areas Estimate the area of the region bounded by the graph of Ζ(π) = xΒ² + 2 and the x-axis on [0, 2] in the following ways.
(a) Divide [0, 2] into n = 4 subintervals and approximate the area of the region using a left Riemann sum. Illustrate the solution geometrically.
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