Problem 59 - Essential Calculus Early Transcendentals

Question

\(59-64=\) Evaluate the integral. $$ \int_{0}^{2}\left(t \mathbf{i}-t^{3} \mathbf{j}+3 t^{5} \mathbf{k}\right) d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( 2\mathbf{i} - 4\mathbf{j} + 32\mathbf{k} \).

Step 1: Identify Components of the Vector Function

The vector function is given as \( t \mathbf{i} - t^3 \mathbf{j} + 3t^5 \mathbf{k} \). We need to integrate each component separately over the interval from 0 to 2.

Step 2: Integrate the \( \mathbf{i} \) Component

The \( \mathbf{i} \) component is \( t \). Integrate this with respect to \( t \): \[ \int_{0}^{2} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = 2. \]

Step 3: Integrate the \( \mathbf{j} \) Component

The \( \mathbf{j} \) component is \( -t^3 \). Integrate this:\[ \int_{0}^{2} -t^3 \, dt = -\left[ \frac{t^4}{4} \right]_{0}^{2} = -\left( \frac{2^4}{4} - \frac{0^4}{4} \right) = -4. \]

Step 4: Integrate the \( \mathbf{k} \) Component

The \( \mathbf{k} \) component is \( 3t^5 \). Integrate this:\[ \int_{0}^{2} 3t^5 \, dt = 3\left[ \frac{t^6}{6} \right]_{0}^{2} = 3\left( \frac{2^6}{6} - \frac{0^6}{6} \right) = 32. \]

Step 5: Combine the Results

Combine the results from Steps 2, 3, and 4 to form the final answer as a vector:\[ 2\mathbf{i} - 4\mathbf{j} + 32\mathbf{k}. \]

Key Concepts

Definite Integrals

Definite integrals are an essential part of calculus. They help determine the total accumulation of a quantity, like area under a curve, between given limits. In the context of vector calculus, definite integrals allow us to process entire vector functions over an interval.

• The limits of integration (such as 0 to 2 in our example) denote the start and end of the interval over which you want to evaluate the function.
• The process involves finding an antiderivative and evaluating it at the boundaries defined by these limits.

This technique is useful not only for scalar functions but also when dealing with vector components, as these too can be integrated similarly, treating each component independently. Thus, understanding definite integrals is crucial for analyzing vector calculus problems effectively.

Vector Components

A vector in space can be decomposed into three components, each corresponding to one of the coordinate axes: \( \mathbf{i}\), \( \mathbf{j}\), and \( \mathbf{k}\). Each letter represents a unit vector pointing in the direction of the corresponding axis:

• \( \mathbf{i} \) is aligned with the x-axis.
• \( \mathbf{j} \) is aligned with the y-axis.
• \( \mathbf{k} \) is aligned with the z-axis.

In our exercise, the vector's components \( t \mathbf{i} \), \( -t^3 \mathbf{j} \), and \( 3t^5 \mathbf{k} \) represent functions that are evaluated along the respective axes. Each component can be integrated separately to resolve its total accumulation over the given interval. This separation into components simplifies complex vector calculus operations, allowing us to handle them using basic calculus integration rules for each term independently.

Integration Techniques

Integration techniques are mathematical methods used to find integrals of functions. We can handle each vector component separately by applying basic integration rules to tackle complex integrations in vector calculus. Here’s a brief look at the techniques used:

• The power rule is the most frequently applied technique for finding antiderivatives. It states that \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).
• When dealing with negatives, such as \( -t^3 \), simply apply the power rule as usual and carry the negative sign through the calculation.
• Constant multipliers, like 3 in \( 3t^5 \), can be factored outside the integral, simplifying the calculation by treating the remaining function independently.

By mastering these fundamental integration techniques, you can efficiently resolve vector calculus problems, breaking down complex functions into manageable segments.