/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Let \(f(x)=c,\) where \(c\) is a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 4

Let \(f(x)=c,\) where \(c\) is a positive constant. Explain why an area function of \(f\) is an increasing function.

Short Answer

Expert verified
Question: Show that an area function of a constant function is an increasing function. Answer: To show that the area function of a constant function, \(f(x) = c\), is an increasing function, we first defined the area function, \(A(x)\), as the integral of \(f(x)\) from a fixed point \(x_0\) up to \(x\). We then calculated the area function for \(f(x) = c\) and found that \(A(x) = c(x - x_0)\). Since \(c\) is a positive constant and \(x - x_0\) increases as \(x\) increases, the area function, \(A(x)\), is also an increasing function.

Step by step solution

01

Define the area function

Let us define an area function \(A(x)\) as follows: \(A(x)\) is the area under the curve of \(f(x)\) from a fixed point \(x_0\) up to \(x\), where \(x_0\) is a constant point on the x-axis. In other words, \(A(x) = \int_{x_0}^{x} f(t) dt\).
02

Compute the area function for f(x) = c

We are given that \(f(x) = c\), where \(c\) is a positive constant. We need to compute the area \(A(x)\), so let's compute the integral of the function. $$A(x) = \int_{x_0}^{x} f(t) dt = \int_{x_0}^{x} c dt$$ Since \(c\) is a constant, it can be taken out of the integral: $$A(x) = c\int_{x_0}^{x} dt$$ Now, integrate with respect to \(t\): $$A(x) = c[t]_{x_0}^{x} = c(x - x_0)$$
03

Analyze the behavior of A(x) as x increases

Now we have the area function \(A(x) = c(x - x_0)\). Let's analyze its behavior as \(x\) increases. We know that \(c\) is a positive constant, therefore, as \(x\) increases, \((x - x_0)\) also increases, which implies that the product \(c(x - x_0)\) will increase as well. Hence, the area function \(A(x)\) is an increasing function as it increases as \(x\) increases. This proves that an area function of a constant function \(f(x) = c\) is indeed an increasing function.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

General results Evaluate the following integrals in which the function \(f\) is unspecified. Note \(f^{(p)}\) is the pth derivative of \(f\) and \(f^{p}\) is the pth power of \(f\). Assume \(f\) and its derivatives are continuous for all real numbers. $$\int 2\left(f^{2}(x)+2 f(x)\right) f(x) f^{\prime}(x) d x$$

Show that the Fresnel integral \(S(x)=\int_{0}^{x} \sin \left(t^{2}\right) d t\) satisfies the (differential) equation \(\left(S^{\prime}(x)\right)^{2}+\left(\frac{S^{\prime \prime}(x)}{2 x}\right)^{2}=1\)

Additional integrals Use a change of variables to evaluate the following integrals. $$\int_{2}^{3} \frac{x}{\sqrt[3]{x^{2}-1}} d x$$

What value of \(b>-1\) maximizes the integral $$\int_{-1}^{b} x^{2}(3-x) d x ?$$

\(\text { Simplify the following expressions.}\) $$\frac{d}{d x} \int_{x}^{1} \sqrt{t^{4}+1} d t$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.