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Chapter 5: Problem 24

Evaluate the integrals using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{1 / 2}^{1} \frac{1}{2 x} d x $$

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{2} \ln(2) \).

Step by step solution

01

Identify the Function and Interval

The integral we need to solve is \( \int_{1 / 2}^{1} \frac{1}{2x} \, dx \). This means we need to integrate the function \( f(x) = \frac{1}{2x} \) over the interval \([\frac{1}{2}, 1]\).
02

Find the Antiderivative

To solve the integral, we need to find the antiderivative of \( \frac{1}{2x} \). Notice that \( \frac{1}{2x} \) can be written as \( \frac{1}{2} \cdot \frac{1}{x} \), which simplifies the process. The antiderivative of \( \frac{1}{x} \) is \( \ln|x| \). So, the antiderivative is \( \frac{1}{2} \ln|x| \).
03

Apply the Fundamental Theorem of Calculus

According to Part 1 of the Fundamental Theorem of Calculus, the value of the definite integral \( \int_{a}^{b} f(x) \, dx \) is given by \( F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \). Here, \( F(x) = \frac{1}{2} \ln|x| \).
04

Evaluate the Antiderivative at the Bounds

Substitute the bounds into the antiderivative:1. \( F(1) = \frac{1}{2} \ln|1| = \frac{1}{2} \cdot 0 = 0 \).2. \( F\left( \frac{1}{2} \right) = \frac{1}{2} \ln\left| \frac{1}{2} \right| = \frac{1}{2} \ln\left( \frac{1}{2} \right) \).
05

Calculate the Definite Integral

Now, compute the definite integral: \[ \int_{1/2}^{1} \frac{1}{2x} \, dx = F(1) - F\left( \frac{1}{2} \right) = 0 - \left( \frac{1}{2} \ln\left( \frac{1}{2} \right) \right) \]This simplifies further to:\[ -\frac{1}{2} \ln\left( \frac{1}{2} \right) \]This can also be written as:\[ \frac{1}{2} \ln(2) \] because \( \ln\left( \frac{1}{2} \right) = - \ln(2) \).

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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 calculation that produces a single number, representing the accumulation of quantities over a specified interval. Think of it like adding up small slices to find a total.
  • Purpose: It helps us find areas under curves, total distances, or even quantities like mass when density is variable over a range.
  • Notation: The definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted as \( \int_{a}^{b} f(x) \, dx \).
  • Result: The output is just a number, representing the total accumulation from point \( a \) to point \( b \).
Notice how we used the limits \( \frac{1}{2} \) and \( 1 \) in this problem. The Fundamental Theorem of Calculus tells us how to evaluate these integrals by using an antiderivative, making things much easier!
Antiderivative
An antiderivative is like the opposite of a derivative. While derivatives measure how a function changes, antiderivatives help us "reverse" this process.
  • Finding: To find an antiderivative, you look for a function whose derivative gives you the original function you're interested in.
  • Example: In our case, we needed the antiderivative of \( \frac{1}{2x} \). This function can be rewritten as \( \frac{1}{2} \cdot \frac{1}{x} \), leading us to recognize it as a logarithmic function.
  • Application: The antiderivative is key to evaluating definite integrals, which is why knowing \( \frac{1}{2}\ln|x| \) was crucial in solving our exercise.
Finding correct antiderivatives makes solving integrals straightforward once you apply the upper and lower limits for evaluation.
Logarithmic Function
Logarithmic functions are intimately tied to the process of integration, especially when dealing with reciprocals or fractions.
  • Definition: A logarithmic function with base \( e \) (Euler's number, approximately 2.718) is usually expressed as \( \ln(x) \), called the natural logarithm.
  • Properties: This function has unique properties that make it useful for integration, such as \( \ln(1) = 0 \) and \( \ln(ab) = \ln(a) + \ln(b) \).
  • Relevance: In our exercise, \( \frac{1}{x} \) was the function from which the logarithmic antiderivative was derived. Recognizing this connection simplifies the process of integration.
Understanding the logarithmic properties allows you to manipulate and simplify expressions within integrals, aiding in finding easier solutions.

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Most popular questions from this chapter

Evaluate the definite integral by expressing it in terms of \(u\) and evaluating the resulting integral using a formula from geometry. $$ \int_{0}^{2} x \sqrt{16-x^{4}} d x ; u=x^{2} $$

Evaluate the integrals by any method. $$ \int_{-1}^{4} \frac{x d x}{\sqrt{5+x}} $$

(a) Give a geometric argument to show that $$ \frac{1}{x+1}<\int_{x}^{x+1} \frac{1}{t} d t<\frac{1}{x}, \quad x>0 $$ (b) Use the result in part (a) to prove that \(\frac{1}{x+1}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x}, \quad x>0\) (c) Use the result in part (b) to prove that $$ e^{x /(x+1)}<\left(1+\frac{1}{x}\right)^{x}0 $$ and hence that $$ \lim _{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x}=e $$ (d) Use the result in part (b) to prove that $$ \left(1+\frac{1}{x}\right)^{x}0 $$

Find the limit $$ \lim _{h \rightarrow 0} \frac{1}{h} \int_{x}^{x+h} \ln t d t $$

Use a calculating utility to find the midpoint approximation of the integral using \(n=20\) sub-intervals, and then find the exact value of the integral using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{1}^{3} \frac{1}{x} d x $$
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