/*! 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 38 Find all functions \(f(t)\) with... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 38

Find all functions \(f(t)\) with the following property: $$f^{\prime}(t)=\frac{4}{6+t}$$

Short Answer

Expert verified
\[ f(t) = 4 \ln|6 + t| + C \]

Step by step solution

01

Understand the problem

Determine the function \( f(t) \) such that its derivative is given by \( f^{\prime}(t) = \frac{4}{6+t} \).
02

Integrate both sides

Find \( f(t) \) by integrating \( \frac{4}{6+t} \) with respect to \( t \). Use the formula for the integral of \( \frac{1}{x} \), which is \( \ln|x| \).
03

Perform the integration

Set up the integral: \[ f(t) = \int \frac{4}{6+t} \; dt \]. Using the substitution \( u = 6 + t \), then \( du = dt \). The integral becomes:\[ f(t) = 4 \int \frac{1}{u} \; du \].
04

Simplify the integral

Integrate \( \frac{1}{u} \): \[ f(t) = 4 \ln|u| + C \]. Substitute back \( u = 6 + t \):\[ f(t) = 4 \ln|6 + t| + C \].
05

Conclude the solution

The general solution for \( f(t) \) is\[ f(t) = 4 \ln|6 + t| + C \] where \( C \) is the constant of integration.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

integration techniques
Integration techniques are methods used to find the integral of a function. An integral is the opposite of a derivative. When given a derivative, finding the original function requires integration.
In this problem, we start with the function's derivative: \(f^{\prime}(t) = \frac{4}{6+t}\) and aim to find the function \(f(t)\).
To achieve this, we integrate \(\frac{4}{6+t}\) with respect to \(t\). The integral is set up as:
\[ f(t) = \int \frac{4}{6+t} \; dt \]
Different functions require different integration techniques. Here, substitution makes the process simpler.
natural logarithm
Natural logarithm, denoted as \(\ln(x)\), is the inverse of the exponential function \(e^x\). It's commonly used in integration, especially when integrating functions of the form \(\frac{1}{x}\).
For example, the integral \( \int \frac{1}{x} \, dx = \ln|x| + C \).
In our problem, after substituting \(u = 6 + t\), the integral becomes:
\[ f(t) = 4 \int \frac{1}{u} \, du = 4 \ln|u| + C \]
Substituting back \(u = 6 + t\), we get:
\[ f(t) = 4 \ln|6 + t| + C \]
Understanding natural logarithms helps make sense of these results.
substitution method
The substitution method simplifies integration by changing variables. This method is used when a direct integration is complicated.
We start by identifying a substitution that makes the integral easier. In our exercise, we set: \( u = 6 + t \).
Then, differentiate \( u \): \[ du = dt. \]
Rewrite the integral with \(u\): \[ \int \frac{4}{6+t} \, dt = 4 \int \frac{1}{u} \, du. \]
Now, integrate \( \frac{1}{u} \): \[ 4 \int \frac{1}{u} \, du = 4 \ln|u| + C. \]
Finally, substitute back \( u = 6 + t \): \[ f(t) = 4 \ln|6 + t| + C. \]
This method often makes integration more manageable.

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

Determine \(\Delta x\) and the midpoints of the subintervals formed by partitioning the given interval into \(n\) subintervals. $$1 \leq x \leq 4 ; n=5$$

We show that, as the number of subintervals increases indefinitely, the Riemann sum approximation of the area under the graph of \(f(x)=x^{2}\) from 0 to 1 approaches the value \(\frac{1}{3}\), which is the exact value of the area. Partition the interval \([0,1]\) into \(n\) equal subintervals of length \(\Delta x=1 / n\) each, and let \(x_{1}, x_{2}, \ldots, x_{n}\) denote the right endpoints of the subintervals. Let $$ S_{n}=\left[f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right)\right] \Delta x $$ denote the Riemann sum that estimates the area under the graph of \(f(x)=x^{2}\) on the interval \(0 \leq x \leq 1\). (a) Show that \(S_{n}=\frac{1}{n^{3}}\left(1^{2}+2^{2}+\cdots+n^{2}\right)\). (b) Using the previous exercise, conclude that $$ S_{n}=\frac{n(n+1)(2 n+1)}{6 n^{3}} $$ (c) As \(n\) increases indefinitely, \(S_{n}\) approaches the area under the curve. Show that this area is \(1 / 3\).

Volume of Solids of Revolution Find the volume of the solid of revolution generated by revolving about the \(x\) -axis the region under each of the following curves. \(y=2 x-x^{2}\) from \(x=0\) to \(x=2\)

Suppose that the interval \(0 \leq x \leq 3\) is divided into 100 subintervals of width \(\Delta x=.03 .\) Let \(x_{1}, x_{2}, \ldots, x_{100}\) be points in these subintervals. Suppose that in a particular application we need to estimate the sum $$ \left(3-x_{1}\right)^{2} \Delta x+\left(3-x_{2}\right)^{2} \Delta x+\cdots+\left(3-x_{100}\right)^{2} \Delta x . $$ Show that this sum is close to 9 .

Suppose that money is deposited steadily in a savings account so that $$\$ 16,000$$ is deposited each year. Determine the balance at the end of 4 years if the account pays \(8 \%\) interest compounded continuously.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

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.