Question

Show that if \(c>0\), then the integral of \(1 / x\) from ac to bc \((0

Short answer

Both integrals evaluate to \(\ln \left( \frac{b}{a} \right)\).

Step-by-step solution

Understand the Problem

We need to compare two definite integrals: the integral of \(1/x\) from \(ac\) to \(bc\) and the integral of \(1/x\) from \(a\) to \(b\), proving they are equal given that \(c > 0\) and \(0 < a < b\).

Write Down the Integrals

The two integrals we are comparing are: \(\int_{ac}^{bc} \frac{1}{x} \, dx\) and \(\int_{a}^{b} \frac{1}{x} \, dx\).

Solve the First Integral

Use the antiderivative of \(1/x\), which is \(\ln|x|\), to evaluate: \[ \int_{ac}^{bc} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{ac}^{bc} = \ln|bc| - \ln|ac|. \]

Simplify the Result of the First Integral

Apply the logarithmic property \(\ln A - \ln B = \ln \left(\frac{A}{B}\right)\): \[ \ln|bc| - \ln|ac| = \ln \left( \frac{bc}{ac} \right) = \ln \left( \frac{b}{a} \right). \]

Solve the Second Integral

The evaluation of the second integral is: \[ \int_{a}^{b} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{a}^{b} = \ln|b| - \ln|a|. \]

Confirm the Equivalence of Results

Observe that both results are the same: \[ \ln \left( \frac{b}{a} \right). \] Thus, \(\int_{ac}^{bc} \frac{1}{x} \, dx = \int_{a}^{b} \frac{1}{x} \, dx\). This confirms that the integrals are indeed equal.

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. They allow us to calculate the area under a curve within a specified interval. The definite integral of a function from a point \(a\) to a point \(b\) is expressed as \(\int_{a}^{b} f(x) \, dx\). It gives a numerical value which represents the accumulated quantity over the interval.
For our exercise, we are comparing two definite integrals of the function \(1/x\) over different intervals. These intervals are transformed by multiplying with a constant, \(c>0\). This fascinating property of the definite integral demonstrates that scaling the bounds by the same positive factor does not affect the integral's value.

Logarithmic Properties

Logarithms have some interesting properties that make calculations simpler and more insightful. One important property is the difference of logarithms, \(\ln A - \ln B = \ln \left(\frac{A}{B}\right)\). This means that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator.
In our context, when evaluating the integral \(\int_{ac}^{bc} \frac{1}{x} \, dx\), we use this property to simplify \(\ln|bc| - \ln|ac|\) into \(\ln \left( \frac{b}{a} \right)\). Understanding and using logarithmic properties like this one is essential to solving many calculus problems efficiently.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that represent the accumulation of change. They "undo" the process of differentiation. For the function \(\frac{1}{x}\), the antiderivative is \(\ln|x|\). This means, if you differentiate \(\ln|x|\), you will get back the original function \(\frac{1}{x}\).
In both parts of our comparison, we use the antiderivative \(\ln|x|\) to evaluate the integrals of \(1/x\) over specific intervals. By substituting the upper and lower bounds into the antiderivative, we determine the accumulation of the function between these points. Understanding antiderivatives is crucial for evaluating definite integrals.

Comparison of Integrals

In this exercise, we compared two integrals to prove their equivalence. The integrals \(\int_{ac}^{bc} \frac{1}{x} \, dx\) and \(\int_{a}^{b} \frac{1}{x} \, dx\) both reduce to \(\ln \left( \frac{b}{a} \right)\).
By showing that both integrals produce the same result, we demonstrate a unique calculus property: the integrals are invariant under scaling the interval by a positive factor \(c\).

• This comparison underscores the power of definite integrals and logarithmic properties to simplify complex interval transformations.
• It illustrates how changes in the limits of integration, when applied uniformly, do not affect the integral's outcome.

This exercise highlights the elegance and consistency within the framework of calculus.