Problem 75 - Applied Calculus

Question

State (without carrying them out) two different methods to find \(\int \ln t d t\)

Step-by-Step Solution

Answer

Methods: Integration by Parts, Known Result Differentiation.

Step 1: Integration by Parts

One method to find the integral of \( \ln t \) is by using integration by parts. Integration by parts is a technique that comes from the product rule for differentiation. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] Here, we choose \( u = \ln t \) and \( dv = dt \). Then, \( du = \frac{1}{t} dt \) and \( v = t \). Applying the integration by parts formula gives us the integral of \( \ln t \).

Step 2: Differentiation of a Known Result

Another method involves using a known result and differentiating it to check its correctness: the integral of \( \ln t \) is \( t \ln t - t + C \), where \( C \) is the constant of integration. This result can be verified by differentiating \( t \ln t - t + C \) to see if it yields \( \ln t \). If it does, we have confirmed our solution.

Key Concepts

Integral of Natural Logarithm

The integral of the natural logarithm, specifically finding \( \int \ln t \, dt \), can appear challenging at first. However, there are effective techniques to solve it.
One of the primary methods used for this integration is integration by parts. However, it's important to start by understanding why the natural logarithm itself is significant in integration.
\( \ln t \) is a unique function because while it seems straightforward, its integral doesn't fit neatly into elementary functions. The trick lies in breaking it down into products or using known results, as with many logarithmic integrals.
The integration of natural logarithms often requires creative approaches, such as manipulation into a product, because the direct antiderivative isn't available through simple formula recognition alone.

Differentiation

Differentiation can verify the results of integration. It’s like checking your math homework answers. Essentially, after computing an integral, you can differentiate the result to ensure accuracy.
Take, for example, the integral \( t \ln t - t + C \). Differentiate each part:

• \( \frac{d}{dt}(t \ln t) = \ln t + 1 \)
• \( \frac{d}{dt}(-t) = -1 \)
• \( \frac{d}{dt}(C) = 0 \)

Combining these derivatives gives \( \ln t \), confirming the integral of \( \ln t \) is correct.
Differentiation not only ensures that calculations are correct but also deepens understanding of the relationship between functions and their derivatives.

Product Rule

The product rule is a fundamental concept in calculus and directly feeds into integration by parts. It's crucial for understanding why integration by parts is constructed the way it is.
The product rule for differentiation is \( \frac{d}{dt}(uv) = u'v + uv' \). This explains why in integration by parts, one chooses parts of the integrand strategically.

• By selecting \( u \) and \( dv \), we can apply integration by parts: \( \int u \, dv = uv - \int v \, du \).
• For \( \ln t \), choosing \( u = \ln t \) ensures that \( du = \frac{1}{t}dt \).
• Meanwhile, \( dv = dt \), leading to \( v = t \).

These selections leverage the product rule to make the solution feasible, transforming difficult integrals into manageable problems.

Constant of Integration

In every integration process, we encounter what's known as the constant of integration, represented by \( C \). This constant captures the infinite family of antiderivatives for a function.
Whenever you integrate, there are multiple functions that could have the same derivative. Adding \( C \) ensures that this ambiguity is accounted for.
For example, with \( t \ln t - t + C \), each different value of \( C \) provides a different function with \( \ln t \) as its derivative.
Remember, the constant of integration is essential not just in theory but also in practical applications, such as solving differential equations where initial conditions help identify the specific value of \( C \).

• Regardless of the function you integrate, always include \( C \) unless initial conditions specify otherwise.

It helps in ensuring that the solution is comprehensive and precisely tailored to the context it's being applied in.