Question

Evaluate the following definite integrals. $$\int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2 t \mathbf{k}) d t$$

Short answer

The definite integral of the vector function over the interval [-π, π] is 2i + 2π²k.

Step-by-step solution

Separate the components of the vector

We are given the vector function as $$ \sin t \mathbf{i} + \cos t \mathbf{j} + 2t \mathbf{k} $$ and we have to find the definite integral with respect to t. We can evaluate the integral of each component separately.

Integrate each component

We have three components, so let's integrate each of them with respect to t from -π to π: 1. Integral of sin(t) with respect to t: $$ \int_{-\pi}^{\pi}\sin t \, dt $$ 2. Integral of cos(t) with respect to t: $$ \int_{-\pi}^{\pi}\cos t \, dt $$ 3. Integral of 2t with respect to t: $$ \int_{-\pi}^{\pi}2t \, dt $$

Evaluate the integrals

Now let's evaluate each integral: 1. $$ \int_{-\pi}^{\pi}\sin t \, dt = \big[-\cos t\big]_{-\pi}^{\pi} = 2 $$ 2. $$ \int_{-\pi}^{\pi}\cos t \, dt = \big[\sin t\big]_{-\pi}^{\pi} = 0 $$ 3. $$ \int_{-\pi}^{\pi}2t \, dt = \big[t^2\big]_{-\pi}^{\pi} = (\pi^2-(-\pi^2)) = 2\pi^2 $$

Combine the components

Now that we have evaluated the integrals for each component of the vector, we can combine the results to find the definite integral of the vector function: $$ \int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2t \mathbf{k}) dt = 2\mathbf{i} + 0\mathbf{j} + 2\pi^2\mathbf{k} $$

Final answer

The definite integral of the vector function over the interval [-π, π] is: $$ 2\mathbf{i} + 2\pi^2\mathbf{k} $$

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and vector functions. In simple terms, a vector function can be thought of as a function where each input results in a vector output. These outputs typically have components in multiple directions, such as the 3D space unit vectors often denoted as \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).

When we perform operations like differentiation or integration on vector functions, we handle each component of the vector separately. This means for an expression like \( \sin t \mathbf{i} + \cos t \mathbf{j} + 2t \mathbf{k} \), we have three separate functions to consider, each representing a component along the respective axis direction. These operations are fundamental in fields like physics and engineering, where we often deal with quantities having both magnitude and direction.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometric terms such as sine and cosine. These integrals are quite common in various applications, from physics to engineering, as they model periodic behavior.

Let’s briefly tackle the integration of \( \sin t \) and \( \cos t \) separately:

• The integral \( \int \sin t \, dt = -\cos t + C \), where \(C\) is the constant of integration. When evaluating the definite integral from \(-\pi\) to \(\pi\), this becomes: \([-\cos t]_{-\pi}^{\pi} = 2\).


• The integral \( \int \cos t \, dt = \sin t + C \). For the limits provided, this evaluates to: \([\sin t]_{-\pi}^{\pi} = 0\). The zero occurs due to the symmetry of the sine function over the interval.

Understanding these results leverages the periodic properties of trigonometric functions, which often simplify calculations when the limits cover full periods of \(2\pi\).

Component-Wise Integration

Component-wise integration is an approach used when integrating vector functions. It simplifies the process by allowing each component of the vector to be integrated separately. This is particularly useful in calculus when dealing with more complex vector-valued functions.

For example, consider the vector function \( \sin t \mathbf{i} + \cos t \mathbf{j} + 2t \mathbf{k} \). The strategy here is to integrate each of the components – \( \sin t \), \( \cos t \), and \( 2t \) – independently, then combine the integrated results.

• The first component \( \int \sin t \, dt \) provides the result for the \( \mathbf{i} \) direction.
• The second component \( \int \cos t \, dt \) addresses the \( \mathbf{j} \) direction, which in this case results in zero due to symmetry.
• The third component \( \int 2t \, dt \) evaluates to \( 2\pi^2 \) providing the \( \mathbf{k} \) component.

Once we have these individual results, we simply recombine them into a single vector expression. This method makes handling such problems more manageable and makes accurate calculations feasible, especially when dealing with definite integrals.