Problem 61 - Calculus III

Question

Evaluate the following integrals:\(\int\left(e^{t} \mathbf{i}+\sin t \mathbf{j}+\frac{1}{2 t-1} \mathbf{k}\right) d t\)

Step-by-Step Solution

Answer

The integral is \(e^t \mathbf{i} - \cos t \mathbf{j} + \frac{1}{2} \ln|2t-1| \mathbf{k} + \mathbf{C}\).

Step 1: Recognize the components of the integral

In the given integral, \[\int\left(e^{t} \mathbf{i}+\sin t \mathbf{j}+\frac{1}{2 t-1} \mathbf{k}\right) d t\]we have a vector function with three components: \(e^t\mathbf{i}\), \(\sin t \mathbf{j}\), and \(\frac{1}{2t-1} \mathbf{k}\). We need to integrate each component separately.

Step 2: Integrate the \(i\)-component

The \(i\)-component of the vector function is \(e^t\). The integral of \(e^t\) with respect to \(t\) is itself: \[\int e^t dt = e^t + C_1\]where \(C_1\) is a constant of integration for the \(i\)-component.

Step 3: Integrate the \(j\)-component

The \(j\)-component of the vector function is \(\sin t\). The integral of \(\sin t\) with respect to \(t\) is:\[\int \sin t dt = -\cos t + C_2\]where \(C_2\) is a constant of integration for the \(j\)-component.

Step 4: Integrate the \(k\)-component

The \(k\)-component of the vector function is \(\frac{1}{2t-1}\). This is a standard form and is recognized as the derivative of the natural logarithm function. Therefore, the integral is:\[\int \frac{1}{2t-1} dt = \frac{1}{2} \ln|2t-1| + C_3\]where \(C_3\) is a constant of integration for the \(k\)-component.

Step 5: Combine results to form the integrated vector

Now, combine the integrated components to form the integrated vector:\[\int \left(e^{t} \mathbf{i} + \sin t \mathbf{j} + \frac{1}{2t-1} \mathbf{k}\right) dt = e^t \mathbf{i} - \cos t \mathbf{j} + \frac{1}{2} \ln|2t-1| \mathbf{k} + \mathbf{C}\]where \(\mathbf{C}\) is the vector of constants \(C_1 \mathbf{i} + C_2 \mathbf{j} + C_3 \mathbf{k}\).

Key Concepts

Vector Integration

When we talk about vector integration, we generally mean integrating vector functions over some domain, much like you do with scalar functions. But here, instead of working with individual numbers, we are working with vectors.
This means that each component of the vector gets integrated separately.

• A vector function might be broken down into components, such as a function in the form \( extbf{F}(t) = f(t) \textbf{i} + g(t) \textbf{j} + h(t) \textbf{k} \).
• Each of these functions—\( f(t) \, \ g(t) \, \text{and}\, h(t) \)—is integrated with respect to the variable, here, "t".
• After integrating each part separately, we combine them back into a single vector.

It may initially seem a little intimidating, especially with all the symbols and dimensions to consider, but much of it boils down to performing usual integrations of scalar functions and then merging those results.

Definite and Indefinite Integrals

Integrals are a fundamental concept in calculus, allowing us to find quantities like areas, volumes, and accumulations. But what distinguishes a definite integral from an indefinite one?

• An **indefinite integral** finds a general form of the antiderivative of a function and often includes a constant of integration (like \( C_1, C_2, \text{and} C_3 \) in our case).
• Indefinite integrals do not evaluate what the antiderivative equals at specific points but rather offer a formula that can be used generally.
• A **definite integral**, on the other hand, computes the integral's value over a specific interval \([a, b]\).
• In a definite integral, the constant of integration is not necessary as the evaluation between limits provides a precise numerical result.

In our exercise, you noticed that constants of integration \( C_1, C_2, \text{and} C_3 \) were included because we were dealing with indefinite integrals, illustrating how the process is more general and applicable to a range of situations.

Vector Functions

Vector functions extend the idea of functions in calculus to higher dimensions. Rather than having an output of a single number for each input, a vector function lets you have a vector as its output.

• A **vector function** often takes the form \( extbf{F}(t) = f(t) \textbf{i} + g(t) \textbf{j} + h(t) \textbf{k} \).
• This means for every value of \( t \), the function outputs a vector composed of the individual magnitudes along the i, j, and k directions.
• The analysis of vector functions differs slightly from scalar functions as it involves considerations of direction, not just size.

In practical terms, this is particularly powerful when describing quantities in physics and engineering that inherently have both magnitude and direction, such as velocity or force fields. This added complexity brings about more sophisticated analytical methods but also a richer set of tools to describe the physical world.