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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 51c
Chapter 1, Problem 51c

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

Verified step by step guidance
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Step 1: Understand the problem. We need to find a formula for A(x), which represents the area under the curve y = f(t) from t = 0 to t = x. The function f(t) is piecewise, meaning it has different expressions depending on the value of t.
Step 2: Identify the intervals. The function f(t) is defined as -2t + 8 for t ≤ 3 and 2 for t > 3. Therefore, we need to consider two cases: when 0 ≤ x ≤ 3 and when x > 3.
Step 3: Calculate A(x) for 0 ≤ x ≤ 3. In this interval, f(t) = -2t + 8. The area A(x) is the integral of f(t) from 0 to x. Set up the integral: A(x) = ∫ from 0 to x of (-2t + 8) dt.
Step 4: Calculate A(x) for x > 3. In this interval, we need to consider the area from 0 to 3 using the first part of the function and from 3 to x using the second part. First, find the area from 0 to 3: A(3) = ∫ from 0 to 3 of (-2t + 8) dt. Then, add the area from 3 to x: ∫ from 3 to x of 2 dt.
Step 5: Combine the results. For 0 ≤ x ≤ 3, A(x) is the result of the integral from Step 3. For x > 3, A(x) is the sum of the area from 0 to 3 and the integral from 3 to x from Step 4. This gives us a piecewise formula for A(x).

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Key Concepts

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

Definite Integral

The definite integral of a function over an interval gives the net area between the function's graph and the x-axis. In this context, to find the area function A(x), we need to compute the definite integral of f(t) from 0 to x. This process involves evaluating the integral based on the piecewise definition of the function f(t) provided.
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Definition of the Definite Integral

Piecewise Functions

A piecewise function is defined by different expressions based on the input value. In this case, f(t) has two distinct expressions: one for t ≤ 3 and another for t > 3. Understanding how to handle piecewise functions is crucial for correctly calculating the area A(x) since the formula will change depending on whether x is less than, equal to, or greater than 3.
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Piecewise Functions

Area Under a Curve

The area under a curve represents the accumulation of values of a function over a specified interval. For the function f(t), the area A(x) can be interpreted as the total area from the t-axis to the curve from t=0 to t=x. This concept is fundamental in calculus, as it connects geometric interpretations with integral calculus, allowing us to quantify the area based on the function's behavior.
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Estimating the Area Under a Curve with Right Endpoints & Midpoint
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.


c. Find a formula for A(x)


ƒ(t) =6 <IMAGE>

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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.


G o g o ƒ

Textbook Question

Use shifts and scalings to transform the graph of ƒ(x)=x2ƒ(x)=x^2 into the graph of g. Use a graphing utility to check your work.


g(x)=ƒ(2x−4)g(x) = ƒ(2x - 4)

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Textbook Question

Shifting a graph Use a shift to explain how the graph of ƒ(x)=−x2+8x+9ƒ(x)=\(\sqrt{-x^2+8x+9}\) is obtained from the graph of g(x)=25−x2g(x)=\(\sqrt{25-x^2}\) . Sketch a graph of ƒ.

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Textbook Question

Use shifts and scalings to transform the graph of ƒ(x)=x2ƒ(x)=x^2 into the graph of g. Use a graphing utility to check your work.

g(x)=6ƒ(x−23)+1g(x)=6ƒ(\(\frac{x-2}{3}\))+1