Problem 20 - Applied Calculus

Question

Both exactly [e.g. \(\ln (3 \pi)]\) and numerically [e.g. \(\ln (3 \pi) \approx 2.243]\). $$ \int_{0}^{5} \ln (1+t) d t $$

Step-by-Step Solution

Answer

The integral is exactly \(6 \ln(6) - 5\) and approximately 5.752.

Step 1: Identify the Integral

Recognize the problem as a definite integral: \(\int_{0}^{5} \ln(1+t) \, dt\). This integral requires integration by parts.

Step 2: Choose Functions for Integration by Parts

Recall the formula for integration by parts: \(\int u \, dv = u \cdot v - \int v \, du\). Choose \(u = \ln(1+t)\) (which simplifies upon differentiation) and \(dv = dt\) (which simplifies upon integration).

Step 3: Differentiate and Integrate the Chosen Functions

Differentiate \(u\) to get \(du = \frac{1}{1+t} \, dt\), and integrate \(dv\) to get \(v = t\).

Step 4: Apply Integration by Parts Formula

Substitute into the integration by parts formula: \(\int u \, dv = (\ln(1+t)) \cdot t - \int t \cdot \frac{1}{1+t} \, dt\), simplifying gives: \(t \ln(1+t) - \int \frac{t}{1+t} \, dt\).

Step 5: Simplify the Integral

Simplify \(\int \frac{t}{1+t} \, dt\) by noting that \(\frac{t}{1+t} = 1 - \frac{1}{1+t}\), which splits the integral into: \(\int (1 - \frac{1}{1+t}) \, dt = \int 1 \, dt - \int \frac{1}{1+t} \, dt\).

Step 6: Integrate the Simplified Expression

The two integrals become \(\int 1 \, dt = t\) and \(\int \frac{1}{1+t} \, dt = \ln|1+t|\). Therefore, \(\int \frac{t}{1+t} \, dt = t - \ln(1+t)\).

Step 7: Substitute Back into Main Expression

Substitution gives: \(t \ln(1+t) - (t - \ln(1+t))\), which simplifies to: \(t \ln(1+t) - t + \ln(1+t)\).

Step 8: Apply Limits from 0 to 5

Substitute the limits into the expression: \([5 \ln(6) - 5 + \ln(6)] - [0 \ln(1) - 0 + \ln(1)]\), which further simplifies to \(5 \ln(6) - 5 + \ln(6)\).

Step 9: Calculate Numerically

Numerically evaluate the expression: \(6 \ln(6) - 5\). Calculate \(\ln(6) \approx 1.792\), so \(6 \times 1.792 = 10.752\). Thus, \(10.752 - 5 = 5.752\).

Key Concepts

Definite Integrals

Definite integrals arise in calculus when we want to compute the accumulated value of a function over a certain interval. This process essentially helps us find the total area under a curve, bounded by the x-axis and two vertical lines that represent the limits of integration.
In our example, we are looking at the integral \[\int_{0}^{5} \ln(1+t) \, dt,\]which means we are interested in the total area under the curve of \(y = \ln(1+t)\) from \(t = 0\) to \(t = 5\).

To solve definite integrals, it is crucial to set up the integral correctly and identify the appropriate limits. These limits (in this case, 0 and 5) tell us exactly where the area begins and ends. Calculating a definite integral involves not only finding the antiderivative of the function but also evaluating this antiderivative at the upper and lower limits.

When dealing with definite integrals, the process often includes:

• Finding the antiderivative


• Applying the Fundamental Theorem of Calculus


• Evaluating the resulting expressions at the given limits

Each of these steps plays a critical role in ensuring the accuracy of your final result.

Integration Techniques

Integration techniques are the methods used to evaluate integrals, and in calculus, there is a variety of these techniques, each suited for different kinds of functions. In our example, we utilized a method known as integration by parts.
Integration by parts is particularly useful when the integration of a product of functions is involved. This technique is derived from the product rule of differentiation and is represented by the formula:
\[\int u \, dv = u \, v - \int v \, du.\]

In our exercise, we chose:

• \(u = \ln(1+t)\)
• \(dv = dt\)

With integration by parts, selecting the correct \(u\) and \(dv\) is crucial. Typically, you would choose \(u\) as a function that simplifies upon differentiation, and \(dv\) as a function that is straightforward to integrate.

The goal of integration by parts is to transform an integral into a simpler one that can be evaluated more easily. This often involves rearranging and manipulating the functions within the integral to eventually find the solution.

Calculus Solutions

In calculus, finding solutions to integrals requires a structured approach and understanding of fundamental principles. The steps taken in solving our integral exemplify a common procedure used in calculus problem-solving. This particular approach leveraged integration by parts, an indispensable tool in the integrator’s toolkit.

Following a systematic approach to calculus solutions usually includes:

• Identifying the type of integral and suitable techniques


• Applying a method systematically to simplify the problem


• Evaluating the antiderivative within given bounds (for definite integrals)


• Checking your results for consistency or possible simplification

Incorporating numerical solutions can also be essential, especially when exact calculations aren’t easily attainable, as was demonstrated through our numerical approximation of \(\ln(6)\).

Furthermore, understanding when to use each mathematical tool can help streamline your problem-solving process in calculus. By practicing these steps, you become more adept at tackling complex calculus problems with confidence.