/*! 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 6 Suppose f is continuous, \(f(0)=... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 6

Suppose f is continuous, \(f(0)=0, f(1)=1, f^{\prime}(x)>0\), and \(\int_{0}^{1} f(x) d x=\frac{1}{3}\). Find the value of the integral \(\int_{0}^{1} \mathrm{f}^{-1}(\mathrm{y}) \mathrm{dy}\)

Short Answer

Expert verified
Answer: The value of the integral is \(\frac{2}{3}\).

Step by step solution

01

Apply Inverse Function Properties

Since f is both continuous and increasing (due to \(f^{\prime}(x) > 0\)), it has an inverse function. We can express the function \(\mathrm{f}^{-1}(\mathrm{y})\) as the inverse of the function f, which means \(f(f^{-1}(y))=y\). Also, note that \(f^{-1}(0)=0\) and \(f^{-1}(1)=1\), since \(f(0)=0\) and \(f(1)=1\). So the integral becomes: $$ \int_{0}^{1} \mathrm{f}^{-1}(\mathrm{y}) \mathrm{dy} $$
02

Change of Variables

Now, let's apply a change of variables. Introduce a new variable \(x=f^{-1}(y)\). Then \(y = f(x)\) and also \(dy = f'(x)dx\). Now the integral becomes: $$ \int_{0}^{1} x f'(x) dx $$ Notice that the limits didn't change since \(f^{-1}(0)=0\) and \(f^{-1}(1)=1\).
03

Integration by Parts

To solve the new integral, we can use integration by parts. We let \(u=x\) and \(dv = f'(x)dx\), so that \(du=dx\) and \(v=f(x)\). The formula for integration by parts states: $$ \int u dv = uv - \int v du $$ Applying this formula to our integral, we get: $$ \int_{0}^{1} x f'(x) dx = \left[xf(x)\right]_{0}^{1} - \int_{0}^{1} f(x) dx $$
04

Evaluate the Integral

Now, we can evaluate the terms in the expression: $$ \left[xf(x)\right]_{0}^{1} = (1\cdot f(1)) - (0\cdot f(0)) = 1 - 0 = 1 $$ The right-hand side integral is given as: $$ \int_{0}^{1} f(x) dx = \frac{1}{3} $$
05

Calculate the Final Value

Now we can plug these values back into the expression from Step 3: $$ \int_{0}^{1} x f'(x) dx = 1 - \frac{1}{3} $$ So, the value of the integral is: $$ \int_{0}^{1} \mathrm{f}^{-1}(\mathrm{y}) \mathrm{dy} = \frac{2}{3} $$

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

Show that \(0.78<\int_{0}^{1} \frac{d x}{\sqrt{1+x^{4}}}<0.93\)

Find the sum of the series \(\frac{x^{2}}{1.2}-\frac{x^{3}}{2.3}+\frac{x^{4}}{3.4}-\ldots+(-1)^{n+1} \frac{x^{n+1}}{n(n+1)}+\ldots,|x|<1\)

Given that f satisfies \(|\mathrm{f}(\mathrm{u})-\mathrm{f}(\mathrm{v})| \leq|\mathrm{u}-\mathrm{v}|\) for \(\mathrm{u}\) and \(v\) in \([a, b]\) then prove that (i) \(\mathrm{f}\) is continuous in \([\mathrm{a}, \mathrm{b}]\) and (ii) \(\left|\int_{a}^{b} f(x) d x-(b-a) f(a)\right| \leq \frac{(b-a)^{2}}{2}\).

Let \(\mathrm{a}>0, \mathrm{~b}>0\), and \(\mathrm{f}\) a continuous strictly increasing function with \(\mathrm{f}(0)=0\). Prove that \(a b \leq \int_{0}^{a} f(x) d x+\int_{0}^{b} f^{-1}(x) d x\) Prove, moreover, that equality occurs if and on ly if \(\mathrm{b}=\mathrm{f}(\mathrm{a})\).

Prove the inequalities: (i) \(\int_{1}^{3} \sqrt{x^{4}+1} d x \geq \frac{26}{3}\)(iii) \(\frac{1}{17} \leq \int_{1}^{2} \frac{1}{1+x^{4}} \mathrm{dx} \leq \frac{7}{24}\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.