Problem 346 - Calculus Volume 1

Question

Write an integral to express the area under the graph of \(y=\frac{1}{t}\) from \(t=1\) to \(e^{x}\) and evaluate the integral.

Step-by-Step Solution

Answer

The integral evaluates to \( x \).

Step 1: Identify the Integral

The problem requires finding the area under the curve of the function \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = e^x \). The integral to represent this area can be written as \( \int_{1}^{e^x} \frac{1}{t} \, dt \).

Step 2: Determine the Antiderivative

The antiderivative of \( \frac{1}{t} \) is the natural logarithm function, \( \ln|t| \). Thus, the antiderivative of \( \frac{1}{t} \) is \( \ln t \).

Step 3: Evaluate the Definite Integral

Evaluate the definite integral by applying the fundamental theorem of calculus. This initial setup is \( \left[ \ln t \right]_{1}^{e^x} = \ln(e^x) - \ln(1) \).

Step 4: Simplify the Expression

Simplify \( \ln(e^x) - \ln(1) \). Since \( \ln(1) = 0 \), the expression becomes \( \ln(e^x) - 0 = x \ln e = x \cdot 1 = x \), as the natural logarithm of \( e \) is 1.

Key Concepts

Natural Logarithm

The natural logarithm, often written as \( \ln \), is a mathematical function that is used to describe growth processes, among other things. It is the inverse of the exponential function with base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. The formula \( y = \ln(x) \) represents a curve that grows slowly and becomes steeper as \( x \) increases. Understanding the natural logarithm is essential in calculus because it shows up frequently in integrals and derivatives. In our problem, the function \( y = \frac{1}{t} \) has an antiderivative of \( \ln|t| \). This relationship is crucial for determining the area under the curve from one point to another, as it allows us to use the logarithm to calculate integrals. The natural logarithm gives us a means to measure or represent the area under this kind of hyperbolic curve.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation with integration, providing a coherent framework for understanding these two central operations in calculus. It asserts that if a function is continuous on a closed interval and \( F \) is an antiderivative of \( f \), then the integral of \( f \) over this interval can be calculated using \( F \). This theorem has two main parts:

• The first part states that if \( f \) is continuous over \([a, b]\), then the function \( F \) defined by the integral from \( a \) to \( x \) of \( f(t) \) is differentiable on \((a, b)\), and \( F'(x) = f(x) \).
• The second part states that if \( F \) is an antiderivative of \( f \) on \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

In our exercise, the Fundamental Theorem of Calculus helps us find the definite integral of \( \frac{1}{t} \) from 1 to \( e^x \). By applying this theorem, we evaluate the antiderivative \( \ln t \) at these bounds to find \( \ln(e^x) - \ln(1) \), simplifying to \( x \).

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that reverse differentiation. In simple terms, if you have a function \( f(x) \), an antiderivative is a function \( F(x) \) whose derivative brings back \( f(x) \). This reversal is fundamental to solving problems involving areas under curves.For the function \( f(x) = \frac{1}{x} \), its antiderivative is \( F(x) = \ln|x| \). This relationship is highly useful because it allows us to evaluate definite integrals easily. The constant of integration is typically unnecessary in definite integrals because it cancels out.In our case, we start with \( y = \frac{1}{t} \) and need the antiderivative to proceed with the integration. Recognizing \( \ln|t| \) as this antiderivative lets us apply the limits of integration from 1 to \( e^x \). This enables us to solve the definite integral, simplifying the process and eventually yielding the solution \( x \) for the area under the curve.