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Chapter 5: Problem 36

Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{1}^{4}(1 / x) d x $$

Short Answer

Expert verified
The integral evaluates to \( \ln 4 \).

Step by step solution

01

Identify the Integrand

The integrand given in the problem is \( \frac{1}{x} \). We need to find its antiderivative, which will help us evaluate the definite integral.
02

Determine the Antiderivative

The antiderivative of \( \frac{1}{x} \) is the natural logarithm function, \( \ln|x| \). Thus, the indefinite integral of \( \frac{1}{x} \) is \( \ln|x| + C \), where \( C \) is the constant of integration.
03

Apply Theorem 3

According to Theorem 3 (Fundamental Theorem of Calculus), if \( F(x) \) is an antiderivative of \( f(x) \) on \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( F(x) = \ln|x| \), \( a = 1 \), and \( b = 4 \).
04

Evaluate the Antiderivative at the Bounds

Substitute \( b = 4 \) and \( a = 1 \) into \( F(x) \): \[ F(4) = \ln|4| = \ln 4 \] \[ F(1) = \ln|1| = \ln 1 = 0 \].
05

Calculate the Definite Integral

Apply the results from Step 4 to Theorem 3: \[ \int_{1}^{4} \frac{1}{x} \, dx = \ln 4 - 0 = \ln 4 \]. This represents the evaluated result of the definite integral.

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Key Concepts

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

Antiderivative
Finding the antiderivative is like uncovering the original function from its derivative. It's a reverse process. In the particular case of the function \( \frac{1}{x} \), its antiderivative is the natural logarithm function, noted as \( \ln|x| \). This means when you differentiate \( \ln|x| \), you get back \( \frac{1}{x} \).

When calculating an indefinite integral, we find the antiderivative and add an arbitrary constant \( C \). So, the integral of \( \frac{1}{x} \) is:
  • \( \int \frac{1}{x} \, dx = \ln|x| + C \)
This constant is crucial in indefinite integrals because there can be many functions with the same derivative. However, when dealing with definite integrals, this constant cancels out when evaluating the limits.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is like a bridge connecting differentiation and integration. It states that if you know the antiderivative \( F(x) \) of a continuous function \( f(x) \) over an interval, you can find the definite integral of \( f(x) \) over that interval. This is expressed mathematically as:

  • \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)
For our integral \( \int_{1}^{4} \frac{1}{x} \, dx \), we use \( F(x) = \ln|x| \) because it's the antiderivative of \( \frac{1}{x} \).

Then, by substituting the bounds, \( a = 1 \) and \( b = 4 \), we find:
  • \( \int_{1}^{4} \frac{1}{x} \, dx = \ln 4 - \ln 1 \)
  • Since \( \ln 1 = 0 \), the result is \( \ln 4 \)
This shows how the theorem makes evaluating definite integrals straightforward once the antiderivative is known.
Natural Logarithm
The natural logarithm, denoted \( \ln(x) \), is a logarithm with the base \( e \), where \( e \approx 2.718 \). It's commonly found in calculus, particularly in solving integrals. In our integral \( \int \frac{1}{x} \, dx \), the natural logarithm appears as the antiderivative.

The natural logarithm has important properties that make it powerful:
  • \( \ln(e) = 1 \)
  • \( \ln(1) = 0 \)
  • The domain is \( x > 0 \) for \( \ln(x) \)
These properties help simplify many calculus problems. In evaluating \( \int_{1}^{4} \frac{1}{x} \, dx \), we directly use \( \ln 1 = 0 \) to simplify our calculation, leading to the final result \( \ln 4 \). This demonstrates how the natural logarithm is not just a mathematical concept, but a handy tool for solving integrals.

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Most popular questions from this chapter

A function \(f,\) an interval \(I,\) and an even integer \(N\) are given. Approximate the integral of \(f\) over \(I\) by partitioning \(I\) into \(N\) equal length subintervals and using the Midpoint Rule, the Trapezoidal Rule, and then Simpson's Rule. $$ f(x)=x^{3} \quad I=[-1 / 2,7 / 2], N=4 $$

A sum of integrals of the form \(\int_{a}^{b} f(x) d x\) is given. Express the sum as a single integral of form \(\int_{c}^{d} g(y) d y\). $$ \int_{-2}^{0} \sqrt{x+2} d x+\int_{0}^{2}(\sqrt{x+2}-x) d x $$

The graphs of \(y=f(x)\) and \(y=g(x)\) intersect in more than two points. Find the total area of the regions that are bounded above and below by the graphs of \(f\) and \(g\). $$ f(x)=x^{3}+x^{2} \quad g(x)=x^{2}+x $$

Calculate the area of the region between the pair of curves. $$ x=y^{2}+6 \quad x=-y^{2}+14 $$

Treat the \(y\) variable as the independent variable and the \(x\) variable as the dependent variable. By integrating with respect to \(y,\) calculate the area of the region that is described. The region between the curves \(y=x\) and \(x=y^{2}-2\)
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