Problem 33 - Advanced Engineering Mathematics

Question

Evaluate the given integral. $$ \int_{-1}^{2}\left(t \mathbf{i}+3 t^{2} \mathbf{j}+4 t^{3} \mathbf{k}\right) d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{3}{2} \mathbf{i} + 9 \mathbf{j} + 15 \mathbf{k} \).

Step 1: Break Down the Integral

The given integral \( \int_{-1}^{2} \left(t \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k}\right) dt \) represents a vector integral with three components: \( t \mathbf{i} \), \( 3t^2 \mathbf{j} \), and \( 4t^3 \mathbf{k} \). This means we need to evaluate each component separately and then combine the results to obtain the final vector.

Step 2: Integrate Each Component

First, integrate each component separately with respect to \( t \):- \( \int_{-1}^{2} t \, dt \) for the \( \mathbf{i} \) component.- \( \int_{-1}^{2} 3t^2 \, dt \) for the \( \mathbf{j} \) component.- \( \int_{-1}^{2} 4t^3 \, dt \) for the \( \mathbf{k} \) component.

Step 3: Compute the \( \mathbf{i} \) Component

The integral \( \int t \, dt = \frac{1}{2}t^2 \) evaluated from \(-1\) to \(2\) is:\[ \left[ \frac{1}{2}t^2 \right]_{-1}^{2} = \frac{1}{2}(2)^2 - \frac{1}{2}(-1)^2 = 2 - \frac{1}{2} = \frac{3}{2} \].Thus, the \( \mathbf{i} \) component is \( \frac{3}{2} \mathbf{i} \).

Step 4: Compute the \( \mathbf{j} \) Component

The integral \( \int 3t^2 \, dt = t^3 \) evaluated from \(-1\) to \(2\) is:\[ \left[ t^3 \right]_{-1}^{2} = (2)^3 - (-1)^3 = 8 - (-1) = 9 \].Thus, the \( \mathbf{j} \) component is \( 9 \mathbf{j} \).

Step 5: Compute the \( \mathbf{k} \) Component

The integral \( \int 4t^3 \, dt = t^4 \) evaluated from \(-1\) to \(2\) is:\[ \left[ t^4 \right]_{-1}^{2} = (2)^4 - (-1)^4 = 16 - 1 = 15 \].Thus, the \( \mathbf{k} \) component is \( 15 \mathbf{k} \).

Step 6: Combine the Components

Combine the calculated components from Steps 3, 4, and 5 to form the final vector result:\( \frac{3}{2} \mathbf{i} + 9 \mathbf{j} + 15 \mathbf{k} \).

Key Concepts

Vector Integration

Vector integration is a fundamental concept in vector calculus. It involves integrating vector functions over a specific interval. Each vector function is a combination of multiple components, each associated with a different unit vector like \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \). To perform a vector integration, you tackle each component separately and integrate them like you would with scalar functions.

In our exercise, we have the vector function \( t \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \). This directly translates to a problem of finding three separate integrals, one for each vector component, over the interval \(-1\) to \(2\). Once calculated, these individual integrals are combined back together to provide the solution as a vector result.

Learning vector integration is essential for fields like physics and engineering, where real-world quantities often translate into vector functions. Understanding how to integrate these enables solving complex real-world problems.

Component-wise Integration

Component-wise integration is an approach where a vector integral is broken down into its individual components. You handle each part separately as if dealing with simple one-variable calculus, making it much easier to solve.

To illustrate, take the integral \( \int_{-1}^{2}(t \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k}) dt \). Here:

• The \( \mathbf{i} \) component is \( t \), so you compute \( \int_{-1}^{2} t \, dt \).
• The \( \mathbf{j} \) component is \( 3t^2 \), leading to \( \int_{-1}^{2} 3t^2 \, dt \).
• For the \( \mathbf{k} \) component, which is \( 4t^3 \), calculate \( \int_{-1}^{2} 4t^3 \, dt \).

After solving these separate integrals, you combine their results. Each solution will give the magnitude along its respective direction, forming the full vector result.

Component-wise integration not only simplifies solving vector calculus problems but also builds a foundation for tackling multi-variable calculus, helping you master solving multidimensional problems one step at a time.

Definite Integrals

Definite integrals are highly useful in calculus for finding the exact area under a curve between two points. For vector integrals, they help in calculating the total contribution of each vector component across a given interval.

In the task, each component of the vector is integrated using definite integrals over \(-1\) to \(2\). For example,

• For \( t \mathbf{i} \), compute \( \int_{-1}^{2} t \, dt \), evaluating it within \(-1\) and \(2\).
• The \( 3t^2 \mathbf{j} \) uses \( \int_{-1}^{2} 3t^2 \, dt \), and
• The \( 4t^3 \mathbf{k} \) requires \( \int_{-1}^{2} 4t^3 \, dt \).

Calculating these will yield specific values, which represent the exact magnitude along each vector direction over the interval.

Mastering definite integrals paves the way for understanding concepts like flux and work done by a force field. It’s a crucial building block for tackling more advanced topics in calculus.