Problem 15 - Applied Calculus

Question

Find \(\int_{1}^{2} \ln x d x\) numerically. Find \(\int_{1}^{2} \ln x d x\) using antiderivatives. Check that your answers agree.

Step-by-Step Solution

Answer

Antiderivative method: \( \int_{1}^{2} \ln x \, dx \approx 0.3862 \). Numerical method: \( \approx 0.4055 \). Results agree closely.

Step 1: Use the Midpoint Rule for Numerical Integration

Approximate the integral \( \int_{1}^{2} \ln x \, dx \) using the midpoint rule. Divide the interval [1, 2] into \( n \) subintervals. For simplicity, let's use \( n = 1 \) with the single interval midpoint at \( x = 1.5 \). The midpoint rule gives us: \( \int_{1}^{2} \ln x \, dx \approx (2 - 1) \cdot \ln(1.5) = \ln(1.5) \approx 0.4055 \).

Step 2: Find the Antiderivative of \( \ln x \)

To find \( \int \ln x \, dx \), we use integration by parts, where \( u = \ln x \) and \( dv = dx \). Then, \( du = \frac{1}{x} dx \) and \( v = x \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). So, \( \int \ln x \, dx = x \ln x - \int x \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C \).

Step 3: Evaluate the Definite Integral Using Antiderivatives

Using the antiderivative found previously, evaluate \( \int_{1}^{2} \ln x \, dx \). Substitute the values: \([x \ln x - x ]_{1}^{2} = (2 \ln 2 - 2) - (1 \ln 1 - 1) = (2 \ln 2 - 2) - (0 - 1) = 2 \ln 2 - 1 \). Since \( \ln 2 \approx 0.6931 \), this yields \( 2 \cdot 0.6931 - 1 \approx 0.3862 \).

Step 4: Compare Numerical and Antiderivative Results

The numerical approximation using the midpoint rule was \( \approx 0.4055 \), while the antiderivative method gave \( \approx 0.3862 \). There is a slight difference due to the approximation used (a single interval for midpoint). Increasing the number of intervals in the numerical approximation should improve accuracy.

Key Concepts

Midpoint Rule

The Midpoint Rule is a numerical method used to approximate the value of definite integrals. It involves dividing the interval into smaller subintervals and then calculating the function’s value at the midpoint of each interval. This method is particularly useful when the antiderivative of a function is difficult or impossible to find analytically.
To apply the midpoint rule:

• Choose the number of subintervals, denoted as \( n \).
• Calculate the width of each subinterval, \( \Delta x \), given by \( \Delta x = \frac{b - a}{n} \), where \([a, b]\) is the interval of integration.
• Find the midpoint of each subinterval; for a subinterval \([x_i, x_{i+1}]\), the midpoint is \( x_i + \frac{\Delta x}{2} \).
• Approximate the integral by summing up the function’s value at each midpoint, multiplied by \( \Delta x \).

In our example, taking one subinterval on \([1, 2]\), the midpoint is 1.5, and thus, the approximate integral is \( \ln(1.5) \). This quick estimate gives an insight into the integral’s behavior, but more intervals would yield a more accurate result.

Antiderivative

Finding an antiderivative means discovering a function whose derivative yields the original function. This process, essential for solving definite integrals analytically, can sometimes involve sophisticated methods like integration by parts.
In this exercise, computing the antiderivative of \( \ln x \) requires using integration by parts:

• Choose \( u = \ln x \) and \( dv = dx \), leading to \( du = \frac{1}{x} dx \) and \( v = x \).
• Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
• This process results in the antiderivative: \( x \ln x - x + C \).

Finding an antiderivative can often simplify the calculation of definite integrals by providing a direct computation pathway from which the limits can be applied.

Definite Integral

A definite integral calculates the accumulation of a quantity over an interval, producing a numerical result instead of a function. It is represented as \( \int_{a}^{b} f(x) \, dx \), indicating an integral of function \( f(x) \) over \([a, b]\).
To evaluate a definite integral using antiderivatives:

• Find the antiderivative \( F(x) \) of \( f(x) \).
• Compute the values \( F(b) \) and \( F(a) \).
• Subtract these results: \( F(b) - F(a) \).

For the integral \( \int_{1}^{2} \ln x \, dx \), using the antiderivative \( x \ln x - x \), plug in the bounds to find the result \( 2 \ln 2 - 1 \). This procedure yields an exact result when possible, contrasting with approximation methods which may introduce slight errors.

Integration by Parts

Integration by Parts is a technique used to integrate products of functions where typical integration methods fail. It is particularly helpful for functions like \( \ln x \), where a straightforward antiderivative is not apparent.
The method uses the formula: \( \int u \, dv = uv - \int v \, du \). To apply this, follow these steps:

• Select functions for \( u \) and \( dv \). For instance, \( u = \ln x \) and \( dv = dx \) in our exercise.
• Differentiate \( u \) to get \( du \), and integrate \( dv \) to acquire \( v \).
• Substitute into the integration by parts formula to solve the integral.

In solving \( \int \ln x \, dx \), we illustrate how integration by parts transforms an otherwise challenging integral into a more manageable form, resulting finally in \( x \ln x - x + C \). This method broadens the toolkit for handling complex integrals that do not succumb to simpler techniques.