Skip to main content
Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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 (𝓍) = βˆ«β‚Λ£ Ζ’(t) dt using the Fundamental Theorem.
Ζ’(𝓍) = cos 𝓍 ; a = 0 , b = Ο€/2 , c = Ο€

Verified step by step guidance
1
Step 1: Recall the Fundamental Theorem of Calculus, which states that if A(𝓍) = βˆ«β‚Λ£ Ζ’(t) dt, then A'(𝓍) = Ζ’(𝓍). This means the derivative of the area function A(𝓍) is equal to the original function Ζ’(𝓍).
Step 2: To find the area function A(𝓍), integrate Ζ’(t) = cos(t) with respect to t from the lower limit a = 0 to the upper limit 𝓍. The integral of cos(t) is sin(t).
Step 3: Apply the definite integral formula: A(𝓍) = βˆ«β‚Λ£ Ζ’(t) dt = [sin(t)]ₐˣ. Substitute the limits of integration into the antiderivative.
Step 4: Substitute the lower limit a = 0 and the upper limit 𝓍 into the expression: A(𝓍) = sin(𝓍) - sin(0). Simplify the result using the fact that sin(0) = 0.
Step 5: The area function A(𝓍) is now expressed as A(𝓍) = sin(𝓍). This function represents the accumulated area under Ζ’(t) = cos(t) from t = 0 to t = 𝓍.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
1m
Was 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 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.
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 bounds, which is crucial for determining the area function A(x) in the given problem.
Recommended video:
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. 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.
Recommended video:
05:06
Finding Area When Bounds Are Not Given
Related Practice
Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume Ζ’, Ζ’', and Ζ’'' are continuous functions for all real numbers.                                                                                                                                                           

                                                                                                                                                                    

(a) βˆ« Ζ’(𝓍) Ζ’'(𝓍) d𝓍 = Β½ (Ζ’(𝓍))Β² + C.

44
views
Textbook Question

Matching functions with area functions Match the functions Ζ’, whose graphs are given in a― d, with the area functions A (𝓍) = βˆ«β‚€Λ£ Ζ’(t) dt, whose graphs are given in A–D.



49
views
Textbook Question

Average value with a parameter Consider the function Ζ’(𝓍) = a𝓍 (1―𝓍) on the interval [0, 1], where a is a positive real number.

(a) Find the average value of Ζ’ as a function of a .

84
views
Textbook Question

{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.

(a) Write the midpoint Riemann sum in sigma notation for an arbitrary value of n.


βˆ«β‚β΄ 2βˆšπ“ d𝓍

81
views
Textbook Question

Sigma notation Evaluate the following expressions.

(a)    10                                                                                                                                                                               

       βˆ‘ ΞΊ                                                                                                                                                                          

       ΞΊ=1                         

64
views
Textbook Question

Approximating displacement The velocity in ft/s of an object moving along a line is given by v = 3tΒ² + 1 on the interval 0 ≀ t ≀ 4, where t is measured in seconds.

(a) Divide the interval [0,4] into n = 4 subintervals, [0,1] , [1.2] , [2,3] , and [3,4]. On each subinterval, assume the object moves at a constant velocity equal to v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0, 4] (see part (a) of the figure)

71
views