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Ch. 1 - Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 1 - FunctionsProblem 49b
Chapter 1, Problem 49b

Area functions Let A(x) be the area of the region bounded by the t -axis and the graph of y=ƒ(t) from t=0 to t=x. Consider the following functions and graphs.




b. Find A(6).




ƒ(t) =6 <IMAGE>

Verified step by step guidance
1
Step 1: Understand the problem. We need to find the area A(x) under the curve y = f(t) from t = 0 to t = x, where f(t) = 6.
Step 2: Set up the integral. Since A(x) is the area under the curve from t = 0 to t = x, we can express it as an integral: A(x) = \( \int_{0}^{x} f(t) \, dt \).
Step 3: Substitute the function into the integral. Since f(t) = 6, the integral becomes: A(x) = \( \int_{0}^{x} 6 \, dt \).
Step 4: Evaluate the integral. The integral of a constant 6 with respect to t is 6t. So, A(x) = \( [6t]_{0}^{x} \).
Step 5: Apply the limits of integration. Substitute the limits into the evaluated integral: A(x) = 6x - 6(0). Therefore, A(x) = 6x. To find A(6), substitute x = 6 into the expression: A(6) = 6(6).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Area Under a Curve

The area under a curve in calculus represents the integral of a function over a specified interval. For a function y = f(t), the area A(x) from t = 0 to t = x is calculated using the definite integral ∫ from 0 to x of f(t) dt. This concept is fundamental in understanding how to compute the total area bounded by the curve and the axes.
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Summary of Curve Sketching Example 2

Definite Integral

A definite integral is a mathematical tool used to calculate the accumulation of quantities, such as area, over a specific interval. It is denoted as ∫ from a to b of f(t) dt, where a and b are the limits of integration. The result of a definite integral is a number that represents the net area between the function and the t-axis over the interval [a, b].
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Introduction to Indefinite Integrals

Function Evaluation

Function evaluation involves substituting a specific value into a function to determine its output. In the context of the problem, evaluating A(6) means calculating the area under the curve from t = 0 to t = 6 for the given function f(t). This process is essential for finding specific values related to the area function A(x).
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Related Practice
Textbook Question

Area functions Let A(x) be the area of the region bounded by the t -axis and the graph of y=ƒ(t) from t=0 to t=x. Consider the following functions and graphs.


a. Find A(2) .

ƒ(t) =6 <IMAGE>

Textbook Question

Area functions Let A(x) be the area of the region bounded by the t -axis and the graph of y=ƒ(t) from t=0 to t=x. Consider the following functions and graphs.


c. Find a formula for A(x)


ƒ(t) =6 <IMAGE>

7
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Textbook Question

More composite functions Let ƒ(x) = | x | , g(x)= x² - 4 , F(x) = √x , G(x) = (1)/(x-2) Determine the following composite functions and give their domains.


ƒ o g

8
views
Textbook Question

Area functions Let A(x) be the area of the region bounded by the t -axis and the graph of y=ƒ(t) from t=0 to t=x. Consider the following functions and graphs.


a. Find A(2) .


ƒ(t) = {-2t+8 if t ≤ 3 ; 2 if t >3 <IMAGE>

Textbook Question

More composite functions Let ƒ(x) = | x | , g(x)= x² - 4 , F(x) = √x , G(x) = (1)/(x-2) Determine the following composite functions and give their domains.


ƒ o G

Textbook Question

More composite functions Let ƒ(x) = | x | , g(x)= x² - 4 , F(x) = √x , G(x) = (1)/(x-2) Determine the following composite functions and give their domains.


G o g o ƒ