Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(b\) so that \(\int_{0}^{b} \ln x d x=0\).

Step-by-Step Solution

Answer

The value of \( b \) is \( e \).

Step 1: Write the Integral Expression

The integral given is \( \int_{0}^{b} \ln x \, dx = 0 \). We need to find the value \( b \) that satisfies this condition.

Step 2: Determine the Antiderivative

Find the antiderivative of the function \( \ln x \). Using integration by parts, let \( u = \ln x \) and \( dv = dx \). Then, \( du = \frac{1}{x}dx \) and \( v = x \). Integrating by parts, \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C \).

Step 3: Evaluate the Definite Integral

Apply the limits of integration to the antiderivative. We have:\[\int_{0}^{b} \ln x \, dx = \left[x \ln x - x\right]_{0}^{b}.\] Now, substitute the values:\[= (b \ln b - b) - \lim_{a \to 0^+}(a \ln a - a).\] As \( a \to 0^+ \), \( a \ln a \to 0 \) and \( -a \to 0 \), thus the limit approaches 0.

Step 4: Solve the Equation

We want \((b \ln b - b) = 0\). Divide by \(b\) assuming \(b eq 0\), we get:\[b \ln b = b.\]This simplifies to \(\ln b = 1\), which gives \(b = e\) (since the exponential function is the inverse of the natural logarithm).

Step 5: Verify the Solution

To ensure this value satisfies \( \int_{0}^{b} \ln x \, dx = 0 \), substitute \( b = e \) into the integral expression:\[\int_{0}^{e} \ln x \, dx = \left[ x \ln x - x \right]_{0}^{e}\]\[= (e \ln e - e) - \lim_{a \to 0^+}(a \ln a - a)\]\[= (e \cdot 1 - e) - 0 = 0 \]Thus, the calculation confirms \( b = e \) satisfies the integral condition.

Key Concepts

Integration by parts

When dealing with the integral of a product of two functions, integration by parts is a very useful technique. It's like the product rule for differentiation, only in reverse. This method helps break down the integral into more manageable pieces, which can sometimes simplify the process significantly. The formula for integration by parts is:

• \(\int u \, dv = uv - \int v \, du\)

where you select parts of the integrand to be \( u \) and \( dv \).
In our exercise, to find the integral of \( \ln x \), we can choose:
- \( u = \ln x \) and \( dv = dx \)

This means \( du = \frac{1}{x} dx \) and \( v = x \).
Then, using the integration by parts formula becomes:
\( \int \ln x \ dx = x \ln x - \int x \cdot \frac{1}{x} \, dx\).
Simplifying further gives \( x \ln x - x + C \) as the antiderivative.

Antiderivative

An antiderivative is a function whose derivative is the given function. In integrals, finding the antiderivative is useful for evaluating definite integrals. For a function \( f(x) \), its antiderivative is often represented by a capital letter such as \( F(x) \).

• The process of finding an antiderivative is called integration.

In our example, we found the antiderivative of the natural logarithm \( \ln x \) to be \( x \ln x - x + C \).
When finding definite integrals from \( a \) to \( b \), we evaluate \( F(b) \) and \( F(a) \), and subtract: \( F(b) - F(a) \). This gives us the area under the curve of the function between these points.

Natural logarithm

The natural logarithm, denoted by \( \ln x \), is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. It is particularly handy in calculus due to its simple derivative and antiderivative characteristics.

• \( \frac{d}{dx}(\ln x) = \frac{1}{x} \)
• \( \int \ln x \, dx = x \ln x - x + C \)

Logarithms play a critical role in solving equations involving growth and decay processes. In this problem, the solution involves finding \( b \) such that the definite integral is zero.
The equation \( b \ln b - b = 0 \) simplifies to finding \( \ln b = 1 \), leading to \( b = e \), because the inverse of the natural logarithm is the exponential function.