Problem 18 - Applied Calculus

Question

Evaluate the integrals in both exactly [e.g. \(\ln (3 \pi)] \text { and numerically [e.g. } \ln (3 \pi) \approx 2.243]\). $$\int_{0}^{5} \ln (1+t) d t$$

Step-by-Step Solution

Answer

Exact: \(6\ln(6) - 5\); Numerical: \(\approx 5.75056\).

Step 1: Understanding the Integral

We need to evaluate the integral \( \int_{0}^{5} \ln(1+t) \, dt \). This means finding the area under the curve of the function \( \ln(1+t) \) from \( t = 0 \) to \( t = 5 \). We will solve this exactly using integration techniques and then estimate it numerically.

Step 2: Integration by Parts

To evaluate \( \int_{0}^{5} \ln(1+t) \, dt \) exactly, we will use integration by parts. Let \( u = \ln(1+t) \) and \( dv = dt \). Then, \( du = \frac{1}{1+t} \, dt \) and \( v = t \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 3: Applying Integration by Parts

Substitute into the integration by parts formula: \[ \int \ln(1+t) \, dt = t\ln(1+t) - \int t \cdot \frac{1}{1+t} \, dt. \]This simplifies to \[ t\ln(1+t) - \int \frac{t}{1+t} \, dt. \]

Step 4: Simplifying the Integral

The term \( \int \frac{t}{1+t} \, dt \) can be simplified to \( \int (1 - \frac{1}{1+t}) \, dt = \int 1 \, dt - \int \frac{1}{1+t} \, dt \).

Step 5: Evaluating the Simplified Integral

Evaluate \[ \int 1 \, dt - \int \frac{1}{1+t} \, dt = t - \ln \left| 1+t \right| + C. \]

Step 6: Combine Results

Substitute back: \[ \int \ln(1+t) \, dt = t\ln(1+t) - (t - \ln(1+t)) + C = t\ln(1+t) - t + \ln(1+t) + C. \]

Step 7: Evaluate Definite Integral

Apply the integral from 0 to 5:Substitute 5 and 0: \[ \left[ 5\ln(6) - 5 + \ln(6) \right] - \left[ 0 \right] = 5\ln(6) - 5 + \ln(6) = 6\ln(6) - 5. \]

Step 8: Calculating the Exact Value

The exact value of the integral is given by \( 6\ln(6) - 5 \).

Step 9: Approximating Numerically

Calculate the numerical approximation:\( \ln(6) \approx 1.79176 \), so the integral value becomes\[ 6\cdot1.79176 - 5 \approx 5.75056. \]

Key Concepts

Definite Integral

When we talk about a definite integral, we're referring to the calculation of the area under a curve for a specific interval. In the exercise, the definite integral is represented as \( \int_{0}^{5} \ln(1+t) \, dt \), which signifies the area beneath the curve of the function \( \ln(1+t) \) as \( t \) ranges from 0 to 5.
This kind of integral gives us an exact numerical value rather than a function with arbitrary constants. It's particularly useful because it provides concrete solutions to problems like finding distances, areas, and volumes.To solve a definite integral, we usually perform indefinite integration first, which involves finding the antiderivative. Afterward, we apply the fundamental theorem of calculus by evaluating this antiderivative at the upper and lower bounds of the interval, and then subtracting these values.
Remember, the definite integral is heavily used in calculus problems to find real-world quantities, which need a precise numerical answer.

Numerical Approximation

While definite integrals provide exact solutions, sometimes it's more practical to use numerical approximation methods, especially when dealing with complex functions or without analytic solutions. Numerical approximation can efficiently give an estimated value of an integral, suitable for such cases. In our solution, this was done by approximating \( \ln(6) \) as 1.79176 and finding the value of the integral \[ 6 \times 1.79176 - 5 \approx 5.75056. \]

Some common numerical methods include:

• Trapezoidal Rule
• Simpson's Rule
• Monte Carlo Integration

These methods break down the area under the curve into simpler geometric shapes and sum them up to estimate the integral. Although these methods yield results that are not exact, they can be especially useful in computational algorithms and real-world problems where exact computations are cumbersome or impossible due to time and computational constraints.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a fundamental concept in calculus and mathematics as a whole. It refers to the logarithm with base \( e \), where \( e \) is the irrational number approximately equal to 2.71828.

In the given exercise, \( \ln(1+t) \) is part of the function we need to integrate. Natural logarithms are often encountered in various mathematical contexts because they simplify multiplicative processes into additive ones, which can be extremely helpful for solving equations and modeling growth processes found in nature and science.When dealing with integration, the properties of the natural logarithm are beneficial because the derivative of \( \ln(x) \) is \( \frac{1}{x} \). This property frequently appears in integration strategies like integration by parts, as used in the solution to the exercise.
Understanding the behavior of \( \ln(x) \), such as how it increases slowly and its undefined nature at or below zero, is crucial for solving integrals and analyzing functions that include logarithmic components.