Problem 346 - Calculus Volume 1

Question

In the following exercises, verify by differentiation that \(\int \ln x d x=x(\ln x-1)+C, \quad\) then use appropriate changes of variables to compute the integral. 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 is \( \int_{1}^{e^x} \frac{1}{t} \, dt = x \).

Step 1: Perform Differentiation

Differentiate the given integral function: \( F(x) = x(\ln x - 1) + C \). We apply the product rule to the term \( x \ln x \). The product rule states that if \( u = x \) and \( v = \ln x \), then \( uv' + vu' = x \cdot \frac{1}{x} + \ln x \cdot 1 \). Therefore, \( F'(x) = \ln x + (x \cdot 0 - 1) = \ln x - 1 + 1 = \ln x \). This verifies the original integral because the derivative of \( F(x) \) returns \( \ln x \).

Step 2: Set Up the Integral Using Change of Variables

We need to compute an integral to express the area under the curve of \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = e^x \). The integral is \( \int_{1}^{e^x} \frac{1}{t} \, dt \).

Step 3: Evaluate the Integral

The integral of \( \frac{1}{t} \) is \( \ln |t| \). So, we evaluate from 1 to \( e^x \). This yields \( \left[ \ln |t| \right]_{1}^{e^x} = \ln(e^x) - \ln(1) = x - 0 = x \). Thus, the area under the graph from 1 to \( e^x \) is simply \( x \).

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used for differentiating products of two functions. When it comes to integration by parts, understanding the product rule is crucial, because it helps us to differentiate complex products effectively. In our case, we were given a function, \( F(x) = x(\ln x - 1) + C \). To verify our integral, we needed to find the derivative of this function, which involved differentiating the product \( x \ln x \). Here's how it works:

• Identify the functions: Let \( u = x \) and \( v = \ln x \).
• Differentiate each: The derivative of \( u \) is \( 1 \), and the derivative of \( v \) is \( \frac{1}{x} \).
• Apply the product rule: According to the product rule, \( (uv)' = u'v + uv' \). Applying this, we get \( 1 \cdot \ln x + x \cdot \frac{1}{x} \), which simplifies to \( \ln x + 1 \).

This matches the derivative of the original integral, confirming that our integration by parts was done correctly.

Definite Integral

A definite integral is used to calculate the exact area under a curve between two limits. It is denoted as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the boundaries of integration. In the exercise, we were tasked with finding the area under the graph of \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = e^x \). Here's the process:

• Formulate the integral: Set up the definite integral: \( \int_{1}^{e^x} \frac{1}{t} \, dt \).
• Evaluate the integral: The antiderivative of \( \frac{1}{t} \) is \( \ln |t| \). So, evaluate this from \( t = 1 \) to \( t = e^x \).
• Compute the result: The result is \( \ln(e^x) - \ln(1) \). Simplifying this gives us \( x \), since \( \ln(e^x) = x \) and \( \ln(1) = 0 \).

Thus, the definite integral confirms that the area under this curve from \( t = 1 \) to \( t = e^x \) is indeed \( x \). Definite integrals provide a powerful way to link geometry and calculus.

Change of Variables

The change of variables technique, also known as substitution, is helpful for simplifying integrals, especially when limits or functions are complex. In our exercise, using proper change of variables made evaluating the integral straightforward. Here's how the concept plays out:

• Identify the variable change: Our function, \( y = \frac{1}{t} \), is tricky to integrate directly between the given limits \( t = 1 \) and \( t = e^x \).
• Substitute the new variable: By setting the upper limit as \( t = e^x \), we simplify the scenario. This means our integral \( \int_{1}^{e^x} \frac{1}{t} \, dt \) benefits from the natural logarithm's properties.
• Solve with substitution: Evaluating the antiderivative, \( \ln |t| \), between 1 and \( e^x \), we leveraged this change effectively. The integral becomes \( \ln(e^x) - \ln(1) = x \).

This change of variables is crucial for making integrals more manageable and directly linked to how integration connects with differentiation through the Fundamental Theorem of Calculus.