Question

For the following exercises, evaluate by any method. $$ \int_{1}^{e^{\pi}} \frac{d x}{x}+\int_{-2}^{-1} \frac{d x}{x} $$

Short answer

The evaluated value is \( \pi - \ln(2) \).

Step-by-step solution

Identify the Integrals

The problem is asking us to evaluate two separate integrals: \( \int_{1}^{e^{\pi}} \frac{d x}{x} \) and \( \int_{-2}^{-1} \frac{d x}{x} \). We will handle each integral independently.

Integrate \( \int_{1}^{e^{\pi}} \frac{d x}{x} \)

The antiderivative of \( \frac{1}{x} \) is \( \ln|x| \). So, \( \int \frac{d x}{x} = \ln|x| + C \). We apply this to the first integral from 1 to \( e^{\pi} \).

Apply Limits to First Integral

Calculate: \( \ln|e^{\pi}| - \ln|1| = \ln(e^{\pi}) - \ln(1) \). Simplifying gives \( \ln(e^{\pi}) = \pi \) (since \( \ln(e^{\pi}) = \pi \ln(e) = \pi \times 1 \)), thus the result is \( \pi \).

Check the Second Integral for Validity

The function \( \frac{1}{x} \) is undefined at \( x = 0 \), thus it should be checked whether the integral from \(-2\) to \(-1\) crosses zero. As it does not, this integral is valid to evaluate over the negative interval.

Integrate \( \int_{-2}^{-1} \frac{d x}{x} \)

Since the antiderivative of \( \frac{1}{x} \) is still \( \ln|x| \), we find the definite integral from \(-2\) to \(-1\).

Apply Limits to Second Integral

Calculate: \( \ln|-1| - \ln|-2| = 0 - \ln(2) \). Simplifying, we get \( -\ln(2) \).

Combine Results from Both Integrals

Add the results from the two evaluated integrals: \( \pi + (-\ln(2)) = \pi - \ln(2) \). This is the final result of the combined integral evaluation.

Key concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \) is the Euler's number. The natural logarithm is a fundamental concept in calculus and higher mathematics due to its role in defining exponential growth and decay. When we say \( \ln|x| \), it implies we take the logarithm of the absolute value of \( x \), ensuring the argument is always positive because the logarithm of a negative number is undefined in the real number system.

Natural logarithms have specific properties:

• \( \ln(1) = 0 \): Because \( e^0 = 1 \), the natural log of 1 is always 0.
• \( \ln(e^x) = x \): This property simplifies many expressions, like in tutorials where \( \ln(e^\pi) = \pi \). The reason is the base of the natural logarithm is \( e \).
• Additive and multiplicative relationships: \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(a^b) = b\ln(a) \). These help in evaluating complex integrals quickly.

Understanding these properties allows for easier manipulation of problems involving exponentials and logarithms, as seen in the solution where \( \ln(e^\pi) \) simplified directly to \( \pi \).

Improper Integral

An improper integral involves limits that are either infinite or pass over a point where the integrand is unbounded. The integral \( \int \frac{dx}{x} \), for instance, is improper at \( x = 0 \) because \( \frac{1}{x} \) is undefined at zero. While our problem doesn't directly cross \( x = 0 \), it's necessary to consider if we ever approach it, especially with limits on intervals like from \(-2\) to \(-1\).

Despite being over the negative interval, these boundaries are valid since it doesn't include a point where the function becomes undefined. Here’s how you handle improper integrals:

• Determine the behavior near the problem point to decide if integration is possible.
• Split the interval if necessary to avoid undefined regions, breaking it into integrals that can be handled separately.

Recognizing these aspects can guide you through evaluating and validating the integrals in more complex expressions, ensuring that computations stay on solid ground.

Antiderivative

An antiderivative of a function is basically the reverse process of differentiation. It is a function whose derivative is the original function you started with. In terms of integrals, finding an antiderivative is essentially finding a function that represents the area under the curve of the given function.

For our specific function \( \frac{1}{x} \), its antiderivative is \( \ln|x| \), which was crucial for solving the problem. Here’s what finding an antiderivative involves:

• Identify a function that, when differentiated, returns you to the original function. For \( \frac{1}{x} \), it's \( \ln|x| \).
• When performing definite integration, apply the found antiderivative function to the upper and lower limits to ascertain the enclosed area or the integral's value.
• Know that the antiderivative includes a "\(+ C\)" constant term for indefinite integrals, but this cancels out in definite integral evaluations, affecting only general solutions.

Understanding antiderivatives enriches your ability to dissect more challenging integrals, offering a key method in the toolkit of calculus for solving various real-world problems.