/*! 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 73 Find the area between the curves... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 73

Find the area between the curves \(y=\ln x\) and \(y=\ln 2 x\) from \(x=1\) to \(x=5 .\)

Short Answer

Expert verified
The area between the curves is \(4 \ln 2\).

Step by step solution

01

Identify the Functions and Bounds

We need to find the area between the curves given by the functions \(y = \ln x\) and \(y = \ln 2x\), within the interval \([1, 5]\).
02

Express the Area as an Integral

The area between the two curves from \(x=1\) to \(x=5\) can be expressed as the integral of the difference of the functions: \[\text{Area} = \int_{1}^{5} (\ln (2x) - \ln x) \, dx\].
03

Simplify the Integral

Simplify \(\ln (2x) - \ln x\) using the properties of logarithms: \[\ln (2x) - \ln x = \ln \left(\frac{2x}{x}\right) = \ln 2\]So the integral becomes:\[\text{Area} = \int_{1}^{5} \ln 2 \, dx\].
04

Compute the Integral

Since \(\ln 2\) is a constant, the integral of \(\ln 2\) from 1 to 5 is:\[\text{Area} = \ln 2 \int_{1}^{5} dx = \ln 2 [x]_{1}^{5}\].
05

Evaluate the Definite Integral

Evaluate the expression:\[\text{Area} = \ln 2 \times (5 - 1) = \ln 2 \times 4 = 4 \ln 2\].

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.

Definite Integral
The concept of a definite integral allows us to find the exact area under a curve between two points on the x-axis. In this problem, the definite integral is used to find the area between two curves, specifically from the functions \(y = \ln x\) and \(y = \ln 2x\) between \(x = 1\) and \(x = 5\). To set up the definite integral, we subtract the lower curve from the upper curve and integrate over the specified interval. This gives us the expression \(\int_{1}^{5} (\ln (2x) - \ln x) \, dx\). Breaking down a problem in this way simplifies calculating areas where functions intersect or integrate over different sections.
Logarithmic Functions
Logarithmic functions, such as \(\ln x\), play a crucial role in calculus and represent a natural way to express exponential relationships. The log function \(\ln x\) is the natural logarithm, which uses the base \(e\), an irrational number approximately equal to 2.718. When dealing with areas between curves, as in this problem, logarithmic functions can simplify to constants, as we will see through the use of their properties. By understanding these functions well, one can easily manipulate and simplify integrals involving them, ultimately making the calculation process smoother and more intuitive.
Properties of Logarithms
The properties of logarithms are powerful tools for simplifying expressions like the one you see in this problem. Here, we use the property \(\ln ab = \ln a + \ln b\), and more importantly, \(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\). In this exercise, \(\ln (2x) - \ln x\) is simplified to \(\ln \left(\frac{2x}{x}\right) = \ln 2\). This drastic simplification from a variable expression to a constant, \(\ln 2\), enables the straightforward integration over the defined interval. Knowing how to apply these properties, one can simplify what appears to be complex problems into easier computations, enhancing both understanding and problem-solving efficiency.

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

For problems \(43-46\) use implicit differentiation to find \(\frac{d y}{d x}\) at the given point \(P .\) $$ 3 \tan ^{-1} x+\sin ^{-1} y=\frac{\pi}{4} ; \quad P(1,-1) $$

The linearization of \(2^{x}\) \begin{equation} \begin{array}{l}{\text { a. Find the linearization of } f(x)=2^{x} \text { at } x=0 . \text { Then round its }} \\ \quad{\text { coefficients to two decimal places. }} \\ {\text { b. Graph the linearization and function together for }} \\ \quad {-3 \leq x \leq 3 \text { and }-1 \leq x \leq 1}.\end{array} \end{equation}

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\csc ^{-1}\left(x^{2}+1\right), x>0 $$

Evaluate the integrals in Exercises \(47-70\) $$ \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+(\sin \theta)^{2}} $$

Evaluate the integrals in Exercises \(71-84\) $$ \int_{1 / 2}^{1} \frac{6 d t}{\sqrt{3+4 t-4 t^{2}}} $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

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.