Problem 22 - 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

Both methods show similar results: Numerically it's approximately 0.34657; analytically it's 0.38629. They agree closely.

Step 1: Numerical Integration Method Selection

To integrate numerically, we can use Simpson's Rule or the Trapezoidal Rule. For this example, we'll use the Trapezoidal Rule, which is more straightforward for manual calculation.

Step 2: Apply the Trapezoidal Rule

The Trapezoidal Rule formula is given by \( T_n = \frac{b-a}{2n} \left( f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right) \). With \( n = 1 \), \( a = 1 \), and \( b = 2 \), it becomes \( T_1 = \frac{2-1}{2} \left( \ln(1) + \ln(2) \right) \). Since \( \ln(1) = 0 \), it simplifies to \( T_1 = \frac{1}{2}(0 + \ln(2)) = \frac{1}{2} \ln(2) \approx 0.34657 \).

Step 3: Find the Antiderivative

The antiderivative of \( \ln x \) can be found using integration by parts, where \( u = \ln x \) and \( dv = dx \). This results in \( \int \ln x \, dx = x \ln x - x + C \).

Step 4: Compute the Definite Integral Using the Antiderivative

To evaluate \( \int_{1}^{2} \ln x \, dx \) using the antiderivative, we calculate \([x \ln x - x]_1^2 = (2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 2 + 1 = 2 \ln 2 - 1 \approx 0.38629 \).

Step 5: Compare Numerical and Antiderivative Results

The numerical result using the Trapezoidal Rule was approximately \(0.34657\), while the exact result using antiderivatives was \(0.38629\). The difference is due to the approximation inherent in numerical methods, but they are relatively close, indicating consistency.

Key Concepts

Trapezoidal Rule

The Trapezoidal Rule is a simple numerical method to estimate the definite integral of a function. It involves approximating the region under the curve as a series of trapezoids rather than using more complex shapes. This approach is particularly useful when the actual antiderivative of the function is difficult or impossible to find.

Here's how it works: You divide the interval into small subintervals, then calculate the area of a trapezoid for each subinterval. The formula for the Trapezoidal Rule with one interval is: \[T_1 = \frac{b-a}{2} \left( f(a) + f(b) \right)\]where:

• \(a\) and \(b\) are the endpoints of the interval.
• \(f(a)\) and \(f(b)\) are the values of the function at the endpoints.

In our example, with \(a = 1\) and \(b = 2\), and \(f(x) = \ln x\), only one trapezoid was used, yielding: \(T_1 = \frac{1}{2} \ln 2 \approx 0.34657\). This provides a quick approximation of the integral, though slight differences from the exact result can occur due to the approximation.

Antiderivatives

Antiderivatives are functions that reverse the process of differentiation. If \(F(x)\) is an antiderivative of \(f(x)\), then differentiating \(F(x)\) will yield \(f(x)\). Finding the antiderivative is crucial for determining the exact value of a definite integral. In other words, it's a way of finding areas under curves more precisely than numerical methods such as the Trapezoidal Rule.

For a function like \(\ln x\), its antiderivative can be tricky, but it can be found using methods such as "Integration by Parts." The antiderivative of \(\ln x\) is \(x \ln x - x + C\), where \(C\) is the constant of integration. When computing the definite integral from 1 to 2, we evaluate:\[[x \ln x - x]_1^2 = (2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 1\]This results in approximately \(0.38629\), which is a more accurate representation of the area under the curve from 1 to 2.

Integration by Parts

Integration by Parts is a powerful technique used to find antiderivatives, especially when dealing with products of functions. This rule is derived from the product rule of differentiation and essentially transforms a difficult integral into a simpler one. It's especially useful for functions like \(\ln x\), where direct integration isn't straightforward.

The formula for Integration by Parts is:\[\int u \, dv = uv - \int v \, du\]In our scenario, to find the antiderivative of \(\ln x\), we choose:

• \(u = \ln x\) and thus \(du = \frac{1}{x} dx\).
• \(dv = dx\) and so \(v = x\).

Plugging these into the formula:\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C\]Using Integration by Parts, we have found the antiderivative, leading us to successfully compute the definite integral \(\int_{1}^{2} \ln x \, dx\) to get the exact result.