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Chapter 8: Problem 26

The integrals in Exercises \(1-34\) converge. Evaluate the integrals without using tables. $$\int_{0}^{1}(-\ln x) d x$$

Short Answer

Expert verified
The value of the integral is 1.

Step by step solution

01

Identify the Integral

We are given the integral \( \int_{0}^{1} (-\ln x) \, dx \). Our task is to find the antiderivative of \(-\ln x\) and then evaluate it over the given interval \([0, 1]\).
02

Express the Integral Function

Let's start by considering that \(-\ln x\) can be integrated by recognizing it as \(-1\times\ln x\). Thus, the integral is expressed as \( \int_{0}^{1} -\ln x \, dx \).
03

Integration by Parts Formula

We'll use the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). Set \( u = -\ln x \) and \( dv = dx \). This means \( du = -\frac{1}{x} \, dx \) and \( v = x \).
04

Apply Integration by Parts

Applying integration by parts, we have: \[ \int (-\ln x) \, dx = x(-\ln x) - \int (-1)x(-\frac{1}{x}) \, dx \] This simplifies to:\[ -x\ln x + \int dx = -x\ln x + x \]
05

Evaluate the Definite Integral

We now evaluate \(-x\ln x + x\) from 0 to 1: Plug in 1: \(-1\ln 1 + 1 = 0 + 1 = 1\).Although \(x\ln x\) is undefined at 0, the limit as x approaches 0 of \(-x\ln x + x\) converges to 0. Thus, the result from 0 to 1 is simply 1.

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Key Concepts

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

Definite Integral
A definite integral is a fundamental concept in calculus that allows us to calculate the area under a curve between two points on the x-axis.
It is denoted as \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the function being integrated, \( a \) is the lower limit, and \( b \) is the upper limit.
To compute a definite integral, we find the antiderivative of the function, evaluate it at the upper and lower limits, and subtract the results.
  • The antiderivative of \( f(x) \) is typically represented as \( F(x) \).
  • The value of the definite integral is calculated as \( F(b) - F(a) \).
In this exercise, we are focusing on the integral \( \int_{0}^{1} (-\ln x) \, dx \). By evaluating this definite integral, we determine the total area between the curve of \(-\ln x\) and the x-axis from 0 to 1.
Integration by Parts
Integration by parts is a powerful technique derived from the product rule of differentiation, used to integrate the products of functions.
The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \), acting as a saving grace when other methods fall short.
  • Choose \( u \) and \( dv \) wisely to simplify the integral. Typically, we want \( du \) to be simpler than \( u \).
  • In our example, we set \( u = -\ln x \) and \( dv = dx \).
  • This means \( du = -\frac{1}{x} \, dx \) and \( v = x \).
After applying integration by parts for \( \int -\ln x \, dx \), we work through the steps, transforming it into a more manageable integral. The process resulted in \( -x\ln x + x \), which simplifies the evaluation of the definite integral over the given limits.
Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718.
Logarithms are the inverse operations of exponentiation, meaning \( \ln(e^x) = x \) and \( e^{\ln x} = x \).
  • \( \ln 1 = 0 \) because \( e^0 = 1 \).
  • The natural logarithm helps simplify exponential growth and decay problems.
In this exercise of integrating \(-\ln x\), we take advantage of its properties while applying integration by parts. Integrating functions involving \( \ln x \) often require clever manipulation, like using integration by parts, to transform difficult integrals into simpler, solvable forms.
Antiderivative
An antiderivative is a function whose derivative is the original function we are trying to integrate. It is also known as the indefinite integral.
For a function \( f(x) \), an antiderivative is signified by \( F(x) \) such that \( F'(x) = f(x) \).
  • The process of finding an antiderivative is known as integration.
  • A definite integral is the difference between the antiderivative evaluated at the upper and lower limits.
In our original problem, we sought the antiderivative of \(-\ln x\) to evaluate the definite integral. Through integration by parts, we derived \(-x\ln x + x\) as the antiderivative of \(-\ln x\). Evaluating this function at 1 and 0, then taking their difference, yielded the final answer for the integral.

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Most popular questions from this chapter

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