/*! 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 8 Suppose that \(f\) is continuous... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 8

Suppose that \(f\) is continuous on \([a, b]\), that \(f(x) \geq 0\) for all \(x \in[a, b]\) and that \(\int_{a}^{b} f=0\). Prove that \(f(x)=0\) for all \(x \in[a, b]\).

Short Answer

Expert verified
Hence, the initial assumption that there exists a point \(c\) such that \(f(c) > 0\) must be incorrect. Therefore, \(f(x) = 0\) for all \(x\) in \([a, b]\). This will be the short answer for the question.

Step by step solution

01

Assume the Contrary

To prove the case, we'll start by assuming the contrary: there exists at least one point \(c\) in the interval \([a, b]\) such that \(f(c) > 0\). This means, that the function value at point \(c\) is greater than zero.
02

Apply the Continuity of the Function

Since \(f\) is continuous on \([a, b]\), by the definition of continuity, there should be a small interval around \(c\), say \((c - \delta, c + \delta)\) within \([a, b]\), where \(f(x) > 0\) for all \(x\) in this interval.
03

Consider the Definite Integral Over This Interval

Now, we consider the definite integral of \(f\) over this small interval \((c - \delta, c + \delta)\). Intuitively, this integral represents the area under the graph of \(f\) over this interval. Since \(f(x) > 0\) for all \(x\) in this interval, this area (and thus, this integral) is strictly greater than zero.
04

Draw a Contrast with Contradiction

However, this contradicts the given information that \(\int_{a}^{b} f = 0\), because this integral, representing the area under the graph of \(f\) over the whole interval \([a, b]\), including the small interval \((c - \delta, c + \delta)\), cannot be zero if the integral over the small interval is strictly greater than zero.

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

If \(\alpha(x):=-x\) and \(\omega(x):=x\) and if \(\alpha(x) \leq f(x) \leq \omega(x)\) for all \(x \in[0,1]\), does it follow from the Squeeze Theorem \(7.2 .3\) that \(f \in \mathcal{R}[0,1]\) ?

Show that one has the estimate \(\left|S_{n}(f)-\int_{a}^{b} f(x) d x\right| \leq\left[(b-a)^{2} / 18 n^{2}\right] B_{2}\), where \(B_{2} \geq\) \(\left|f^{\prime \prime}(x)\right|\) for all \(x \in[a, b] .\)

Use the Substitution Theorem \(7.3 .8\) to evaluate the following integrals. (a) \(\int_{0}^{1} t \sqrt{1+t^{2}} d t\), (b) \(\int_{0}^{2} t^{2}\left(1+t^{3}\right)^{-1 / 2} d t=4 / 3\), (c) \(\int_{1}^{4} \frac{\sqrt{1+\sqrt{t}}}{\sqrt{t}} d t\) (d) \(\int_{1}^{4} \frac{\cos \sqrt{t}}{\sqrt{t}} d t=2(\sin 2-\sin 1)\).

Let \(f:[a, b] \rightarrow \mathbb{R}\) be continuous on \([a, b]\) and let \(v:[c, d] \rightarrow \mathbb{R}\) be differentiable on \([c, d]\) with \(v([c, d]) \subseteq[a, b]\). If we define \(G(x):=\int_{a}^{v(x)} f\), show that \(G^{\prime}(x)=f(v(x)) \cdot v^{\prime}(x)\) for all \(x \in[c, d]\)

Approximate the indicated integrals, giving estimates for the error. Use a calculator to obtain a high degree of precision.$$ \int_{0}^{1} \cos \left(x^{2}\right) d x $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

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.