Problem 13 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in problem exactly. $$ \int_{1}^{2} \frac{1}{2 t} d t $$

Step-by-Step Solution

Answer

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

Step 1: Find the Antiderivative

Identify the function to be integrated which is \(\frac{1}{2t}\). The antiderivative of \(\frac{1}{t}\) is \(\ln|t|\). Therefore, the antiderivative of \(\frac{1}{2t}\) is \(\frac{1}{2}\ln|t|\).

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \(F\) is an antiderivative of \(f\) over the interval \([a, b]\), then \(\int_{a}^{b} f(t) \, dt = F(b) - F(a)\).

Step 3: Evaluate the Antiderivative at the Bounds

The antiderivative found is \(F(t) = \frac{1}{2}\ln|t|\). Evaluate it at the bounds: \(F(2)\) and \(F(1)\). \[F(2) = \frac{1}{2}\ln|2|\] \[F(1) = \frac{1}{2}\ln|1|\] Since \(\ln 1 = 0\), we have \(F(1) = 0\).

Step 4: Subtract the Evaluations

Using the results from Step 3, subtract the evaluated lower boundary from the upper boundary: \[F(2) - F(1) = \frac{1}{2}\ln|2| - 0 = \frac{1}{2}\ln 2\].

Step 5: Final Result

The final result of the definite integral is \(\frac{1}{2}\ln 2\).

Key Concepts

Definite Integral

The definite integral is a key concept in calculus, providing the accumulated sum of values of a function over an interval. It is denoted by the integral symbol with limits, such as \ \( \int_{a}^{b} f(t) \, dt \ \). This represents the total area under the curve of the function \( f(t) \) from \( a \) to \( b \).

To calculate a definite integral, you first need to find an antiderivative \( F(t) \) of \( f(t) \). The Fundamental Theorem of Calculus comes into play here. It connects the antiderivative with the definite integral, stating that if \( F(t) \) is an antiderivative of \( f(t) \), then

• \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \)

This theorem simplifies the process significantly, as it only requires evaluating the antiderivative at two points, \( a \) and \( b \), and then finding the difference between these two values. It's a powerful tool for solving a wide variety of problems in calculus.

Antiderivative

An antiderivative is a function whose derivative results in the original function. In other words, if \( F(t) \) is an antiderivative of \( f(t) \), then \( F'(t) = f(t) \). Finding antiderivatives is essential for calculating definite integrals.

To illustrate, consider the function \( \frac{1}{t} \). The natural logarithm function \( \ln|t| \) is its antiderivative because the derivative of \( \ln|t| \) is \( \frac{1}{t} \). When the function is scaled, such as \( \frac{1}{2t} \), its antiderivative becomes \( \frac{1}{2}\ln|t| \).

The process of finding an antiderivative involves applying rules of differentiation in reverse. This skill is essential for students learning calculus, as it simplifies the task of computing areas under curves, among other applications.

Natural Logarithm

The natural logarithm is a specific logarithmic function, denoted as \( \ln t \), which has a unique base \( e \), an important constant approximately equal to 2.71828. It is defined for all positive numbers \( t \) and is an inverse operation to exponentiation with the same base. This function is significant in calculus and science for various reasons.

One key property of the natural logarithm is its derivative, which is \( \frac{1}{t} \). This property makes it a perfect candidate for finding antiderivatives of expressions like \( \frac{1}{t} \). This matches the natural process, where exponentiation and logarithms are inverse operations to each other.

• The derivative of \( \ln t \) is \( \frac{1}{t} \).
• It helps in solving integrals involving functions such as \( \frac{1}{t} \).

In practice, understanding the natural logarithm enables the calculation of many common integrals, which frequently appear in both physical sciences and financial models. Being aware of its properties and behavior will aid students in mastering calculus concepts efficiently.