Problem 34 - Advanced Engineering Mathematics

Question

In Problems, evaluate the given integral. $$ \int_{0}^{4}(\sqrt{2 t+1} \mathbf{i}-\sqrt{t} \mathbf{j}+\sin \pi t \mathbf{k}) d t $$

Step-by-Step Solution

Answer

\(\frac{26}{3}\textbf{i} - \frac{16}{3}\textbf{j}\)

Step 1: Split the Integral into Components

Identify the vector components of the integral, which consists of three separate functions: \[\int_{0}^{4} \sqrt{2t+1} \, dt \textbf{i}, \quad \int_{0}^{4} \sqrt{t} \, dt \textbf{j}, \quad \int_{0}^{4} \sin(\pi t) \, dt \textbf{k}.\] We will evaluate each integral separately.

Step 2: Evaluate the 1st Component Integral

The first integral is \(\int_{0}^{4} \sqrt{2t+1} \, dt\). This involves a substitution where \(u = 2t + 1\), hence \(du = 2 \, dt\) or \(dt = \frac{1}{2} du\). The limits change as follows: When \(t = 0\), \(u = 1\); and when \(t = 4\), \(u = 9\). The integral becomes: \[\frac{1}{2}\int_{1}^{9} \sqrt{u} \, du.\]This results in \[\frac{1}{2} \left. \frac{2}{3} u^{3/2} \right|_{1}^{9} = \frac{1}{3}(9^{3/2} - 1^{3/2}) = \frac{1}{3}(27 - 1) = \frac{26}{3}.\] So, the result for the **i** component is \(\frac{26}{3}\textbf{i}\).

Step 3: Evaluate the 2nd Component Integral

The second integral is \(\int_{0}^{4} \sqrt{t} \, dt\). We use the power rule to integrate: \[\left. \frac{2}{3}t^{3/2} \right|_{0}^{4}.\] Evaluating this gives: \[\frac{2}{3}(4^{3/2} - 0^{3/2}) = \frac{2}{3}(8) = \frac{16}{3}.\] Thus, the result for the **j** component is \(-\frac{16}{3}\textbf{j} \) (notice the negative sign from the original problem statement).

Step 4: Evaluate the 3rd Component Integral

The third integral is \(\int_{0}^{4} \sin(\pi t) \, dt\). By substitution, let \(u = \pi t\) so \(du = \pi \, dt\) or \(dt = \frac{1}{\pi} du\). The limits change accordingly: when \(t = 0\), \(u = 0\); and when \(t = 4\), \(u = 4\pi\). The integral becomes: \[\frac{1}{\pi}\int_{0}^{4\pi} \sin u \, du.\]Integrating gives: \[\frac{1}{\pi} \left. -\cos u \right|_{0}^{4\pi} = \frac{1}{\pi}(-\cos 4\pi + \cos 0) = \frac{1}{\pi}(1 - 1) = 0.\] The result for the **k** component is \(0\textbf{k}\).

Step 5: Combine the Results

Put together the results from all three components to form the final result of the original vector integral. The integrals thus combine to become:\[ \int_{0}^{4}(\sqrt{2t+1} \textbf{i} - \sqrt{t} \textbf{j} + \sin\pi t \textbf{k}) \ dt = \frac{26}{3}\textbf{i} - \frac{16}{3}\textbf{j} + 0\textbf{k}.\]

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, representing the signed area under a curve between two points. These are calculated using integration, which is the reverse process of differentiation. For the integral

• The limits of integration, such as from 0 to 4 in our example, define the range over which we are integrating.
• Definite integrals have not only applications in finding areas but also in physics for calculating quantities like work and mass.

The process starts by finding the antiderivative of the function being integrated, which is a function whose derivative is the original function. Once the antiderivative is determined, we substitute the upper and lower limits and find the difference to get the result. This difference gives us the value of the area or quantity within those bounds.

Vector Integrals

Vector integrals involve integrating vector functions, which have more than one component, typically denoted as \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in three dimensions. Each of these components can be thought of as a separate function that requires its own integration.

• Each component of a vector integral is integrated as a separate function. The integrals from each dimension provide part of the final vector result.
• To solve vector integrals, it typically involves integrating each component of the vector function independently over the same interval.

Once all components have been integrated separately, their results are recombined to give the final vector. This allows us to evaluate the overall effect of the vector function over the specified domain.

Component-wise Integration

Component-wise integration means treating each component of a vector separately when integrating. In the context of a vector integral:

• It allows us to focus on the integral of scalar functions before considering the vector as a whole.
• For example, for a vector with components \( \textbf{i}, \textbf{j}, \textbf{k} \), we evaluate the integrals \( \int f(t) \, dt \mathbf{i}, \int g(t) \, dt \mathbf{j}, \int h(t) \, dt \mathbf{k} \) separately.

This approach simplifies the problem, breaking it into manageable parts. Each scalar integral is solved using standard techniques like substitution or integration by parts. After solving each integral, we recreate the original vector by combining the evaluated components. This results in a complete understanding of the vector integral in all dimensions.