Problem 18 - Calculus Single Variable

Question

In Exercises \(11-20,\) calculate \(F(x)=\int_{a}^{x} f(t) d t\) $$ f(t)=1 / t \quad a=e \quad(0

Step-by-Step Solution

Answer

\(F(x) = \ln x - 1\) for \(x > 0\).

Step 1: Identify the Integral

We need to calculate the integral function \(F(x)\) given by \(F(x) = \int_{a}^{x} f(t) dt\), where \(f(t) = \frac{1}{t}\) and \(a = e\). This implies that we must evaluate the definite integral of \(\frac{1}{t}\) from \(e\) to \(x\).

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \(F(x)\) is defined as \(\int_{a}^{x} f(t) dt\), then \(F'(x) = f(x)\). To find \(F(x)\), we integrate \(f(t) = \frac{1}{t}\) from \(a\) to \(x\). The antiderivative of \(\frac{1}{t}\) is \(\ln|t|\).

Step 3: Evaluate the Definite Integral

Calculate \(F(x) = \left[ \ln |t| \right]_e^x = \ln |x| - \ln |e|\). Since \(|e| = e\) and \(\ln e=1\), this simplifies to: \[ F(x) = \ln |x| - 1 \].

Step 4: Express the Final Result

Our final expression for the integral \(F(x)\) is: \[ F(x) = \ln x - 1 \]. This is valid for \(x > 0\) since \(f(t) = \frac{1}{t}\) requires \(t\) to be non-zero and positive.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus used to determine the accumulated change or area under a curve between two points. The two points, called limits of integration, specify the interval over which we sum the values of a function. In our exercise, we calculate a definite integral of the function \(f(t) = \frac{1}{t}\) from \(t = e\) to \(t = x\).

• The lower limit is \(a = e\).
• The upper limit is \(x\), representing a variable endpoint.

The process involves finding an antiderivative (a function whose derivative is the original function) and then applying the Fundamental Theorem of Calculus. This theorem connects differentiation and integration, allowing us to evaluate the integral by finding the difference in the antiderivative's values at the two limits.
Thus, the definite integral \(F(x) = \int_{e}^{x} \frac{1}{t} dt\) tells us the total area under the curve from \(e\) to \(x\), yielding \(F(x) = \ln |x| - \ln |e|\). Learning how to calculate definite integrals is crucial for understanding real-life phenomena involving change and accumulation.

Antiderivative

An antiderivative is a function whose derivative is the original function. For example, if you differentiate the function \(F(x)\) and obtain a specific function \(f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). In the exercise, you are tasked with finding the antiderivative of \(f(t) = \frac{1}{t}\).

• The antiderivative of \(\frac{1}{t}\) is \(\ln|t| + C\). Here, \(C\) is the constant of integration, but given the context of definite integrals, we find specific values.
• To evaluate \(F(x)\), we substitute \(t\) with the bounds \(e\) and \(x\).

This results in evaluating the expression \([F(t)]_e^x = \ln |x| - \ln |e|\). In simpler terms, you're finding a specific antiderivative that satisfies the accumulation of values within the bounds. Recognizing antiderivatives is crucial for solving integral calculus problems as it allows calculating changes cumulatively over an interval.

Properties of Logarithms

Logarithms are mathematical functions, often simplified using their unique properties, making them essential tools for expressions involving exponential or growth patterns. In our specific problem, properties of logarithms help in simplifying the expression after integration.

One key property of logarithms applied here is:

• \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\)

Using this property, the expression \(\ln |x| - \ln |e|\) simplifies to \(\ln \left( \frac{x}{e} \right)\). However, since \(\ln e = 1\), the term simply reduces to \(\ln x - 1\). This simplification shows how logarithmic properties efficiently simplify expressions and solve integrals involving natural logarithms. Understanding these properties will help you work through logarithmic terms encountered in calculus and beyond, making problem-solving more manageable.