Indefinite Integral Of Vector Function: Step-by-Step Solution

Hey guys! Today, we're diving into the fascinating world of vector calculus to tackle a common problem: finding the indefinite integral of a vector function. Specifically, we'll be working through an example step-by-step to make sure you've got a solid grasp of the process. So, let's jump right in!

Problem Statement

Our mission, should we choose to accept it, is to find the indefinite integral of the vector function:

$\vec{r}(t) = \langle -2t^4 - 4, -5e^{-5t}, 3\sin(-5t) \rangle$

Remember, the indefinite integral is essentially the antiderivative, and we're looking for a function whose derivative gives us the original vector function. It sounds complex, but we'll break it down into manageable pieces.

Understanding Vector Functions

Before we dive into the integration process, it’s crucial to understand what a vector function is. A vector function is a function that maps a real number (in our case, t) to a vector. Think of it as a curve in space, where each value of t corresponds to a point on the curve. The components of the vector function represent the coordinates of the point in terms of t.

In our example, $\vec{r}(t)$ has three components:

• -2t^4 - 4 (the x-component)
• -5e^{-5t} (the y-component)
• 3sin(-5t) (the z-component)

Each of these components is a scalar function of t. Integrating a vector function involves integrating each of these components separately. This is a key concept that simplifies the whole process. Remember, each component is treated individually, making the problem much more approachable.

Step 1: Integrate Each Component

The magic of integrating vector functions lies in its simplicity: we integrate each component separately. Let's break down the integration of each component:

1. Integrating the x-component: -2t^4 - 4

This is a straightforward polynomial integration. We'll use the power rule, which states that the integral of ${t^n}$ is ${\frac{t^{n+1}}{n+1}}$. So,

$\int (-2t^4 - 4) dt = -2 \int t^4 dt - 4 \int dt$

Applying the power rule:

• ${\int t^4 dt = \frac{t^5}{5}}$
• ${\int dt = t}$

Thus, the integral of the x-component is:

${-2 \cdot \frac{t^5}{5} - 4t = -\frac{2}{5}t^5 - 4t}$

Don't forget, we'll add the constant of integration later.

2. Integrating the y-component: -5e^{-5t}

For this component, we're dealing with an exponential function. Recall that the integral of ${e^{kt}}$ is ${\frac{1}{k}e^{kt}}$. Here, we have ${-5e^{-5t}}$, so:

$\int -5e^{-5t} dt = -5 \int e^{-5t} dt$

Applying the integral rule for exponentials:

$\int e^{-5t} dt = \frac{e^{-5t}}{-5}$

Therefore, the integral of the y-component is:

${-5 \cdot \frac{e^{-5t}}{-5} = e^{-5t}}$

3. Integrating the z-component: 3sin(-5t)

Now, let's tackle the trigonometric function. The integral of ${\sin(kt)}$ is ${-\frac{1}{k}\cos(kt)}$. So, for 3sin(-5t):

$\int 3\sin(-5t) dt = 3 \int \sin(-5t) dt$

Applying the integral rule for sine:

$\int \sin(-5t) dt = -\frac{1}{-5}\cos(-5t) = \frac{1}{5}\cos(-5t)$

Thus, the integral of the z-component is:

${3 \cdot \frac{1}{5}\cos(-5t) = \frac{3}{5}\cos(-5t)}$

Step 2: Combine the Components and Add the Constant Vector

Now that we've integrated each component, let's put them back together to form the indefinite integral of the vector function. We have:

• x-component integral: ${-\frac{2}{5}t^5 - 4t}$
• y-component integral: ${e^{-5t}}$
• z-component integral: ${\frac{3}{5}\cos(-5t)}$

So, the vector formed by these integrals is:

$\langle -\frac{2}{5}t^5 - 4t, e^{-5t}, \frac{3}{5}\cos(-5t) \rangle$

But we're not quite done yet! Remember that indefinite integrals always have a constant of integration. Since we're dealing with a vector function, we need to add a constant vector ${\vec{C} = \langle C_1, C_2, C_3 \rangle}$, where ${C_1}$, ${C_2}$, and ${C_3}$ are arbitrary constants.

So, the final indefinite integral is:

$\int \vec{r}(t) dt = \langle -\frac{2}{5}t^5 - 4t + C_1, e^{-5t} + C_2, \frac{3}{5}\cos(-5t) + C_3 \rangle$

However, the problem asks us not to include the constant vector, so the final answer we'll present is:

$\langle -\frac{2}{5}t^5 - 4t, e^{-5t}, \frac{3}{5}\cos(-5t) \rangle$

Key Takeaways

Let's recap the main points we covered:

1. Vector functions are functions that output vectors, with each component being a scalar function of the input variable.
2. To integrate a vector function, integrate each component separately.
3. Remember to add a constant vector when finding the indefinite integral, but in this specific problem, we omitted it as requested.
4. Pay close attention to the rules of integration for polynomials, exponentials, and trigonometric functions.

Common Mistakes to Avoid

• Forgetting the constant of integration: Always remember to add the constant vector when finding indefinite integrals.
• Incorrectly applying integration rules: Make sure you know the integral rules for different types of functions.
• Mixing up integration and differentiation rules: It's easy to confuse the rules for integration and differentiation, so double-check!
• Not integrating each component: Remember, each component of the vector function must be integrated separately.

Practice Makes Perfect

Integrating vector functions might seem tricky at first, but with practice, you'll become a pro! Try working through different examples, varying the complexity of the components. The more you practice, the more comfortable you'll become with the process.

Conclusion

And there you have it! We've successfully found the indefinite integral of the vector function $\vec{r}(t)$. Remember, the key is to break down the problem into smaller, manageable steps and tackle each component individually. Keep practicing, and you'll master vector integration in no time! Keep an eye out for more math tutorials, guys! Happy integrating!