Integrating ∫ 6e^(-3x) Dx: A Step-by-Step Guide

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Integrating ∫ 6e^(-3x) dx: A Step-by-Step Guide

Hey guys! Today, we're diving into the world of calculus to tackle the integral of 6e^(-3x) dx. If you're just starting out with integration or need a refresher, you've come to the right place. We'll break it down step by step, making sure you understand the underlying concepts and the mechanics of solving this integral. So, grab your pencils, and let's get started!

Understanding the Integral

Before we jump into solving the integral, let's understand what we're dealing with. The expression ∫ 6e^(-3x) dx represents the indefinite integral of the function 6e^(-3x) with respect to x. In simpler terms, we're looking for a function whose derivative is 6e^(-3x). This involves reversing the process of differentiation. Integrals are a fundamental concept in calculus with vast applications in physics, engineering, economics, and many other fields. They help us find areas, volumes, and the accumulation of quantities, and are essential tools for solving differential equations and modeling real-world phenomena.

What is Integration?

Integration is basically the reverse process of differentiation. Think of it like this: if differentiation tells you the slope of a curve at any point, integration helps you find the area under the curve. When we talk about the indefinite integral, we’re looking for a family of functions (because adding a constant to a function doesn't change its derivative) whose derivative is the function we started with.

The Importance of the Constant of Integration

One crucial thing to remember is the constant of integration, often denoted as "C." When we find an indefinite integral, we always add "C" because the derivative of a constant is zero. This means there are infinitely many functions that could have the same derivative. For example, the derivative of x^2 + 1 is 2x, but so is the derivative of x^2 + 5 or x^2 - 100. The "C" accounts for all these possibilities. Omitting the constant of integration can lead to errors in problem-solving, especially when dealing with initial value problems or applications requiring specific solutions.

Setting up the Integral

The integral we need to solve is:

∫ 6e^(-3x) dx

The first thing we can do to simplify this is to pull the constant 6 outside the integral. This is a handy property of integrals: the integral of a constant times a function is the constant times the integral of the function. Mathematically, it looks like this:

∫ k * f(x) dx = k * ∫ f(x) dx

Where 'k' is a constant and 'f(x)' is our function. Applying this to our integral, we get:

6 ∫ e^(-3x) dx

This simplifies our problem a bit and makes it easier to focus on the exponential part of the integral.

Using u-Substitution

Now, we need to tackle the integral ∫ e^(-3x) dx. This is where a technique called u-substitution comes in handy. U-substitution is a powerful method for simplifying integrals by changing the variable of integration. It's essentially the reverse of the chain rule in differentiation. The goal is to replace a complex expression within the integral with a single variable, making the integral easier to solve.

Identifying the 'u'

The key to u-substitution is choosing the right expression to be our 'u'. In this case, the exponent -3x looks like a good candidate. So, let's set:

u = -3x

Finding 'du'

Next, we need to find the derivative of u with respect to x, which we call 'du/dx'.

du/dx = -3

Now, we want to isolate 'dx' to substitute it back into our integral. We can do this by multiplying both sides by 'dx' and dividing by -3:

dx = du / -3

or

dx = -1/3 du

Substituting into the Integral

Now we have everything we need to substitute back into our integral. We replace -3x with u and dx with -1/3 du:

6 ∫ e^(-3x) dx becomes 6 ∫ e^(u) (-1/3) du

Let's pull the constant -1/3 outside the integral as well:

6 * (-1/3) ∫ e^(u) du

This simplifies to:

-2 ∫ e^(u) du

Integrating the Simplified Expression

Now we have a much simpler integral to solve: ∫ e^(u) du. This is a basic integral that you'll often encounter in calculus.

The integral of e^(u) with respect to u is simply e^(u). So, we have:

-2 ∫ e^(u) du = -2e^(u) + C

Don't forget the constant of integration, 'C'! It's super important.

Substituting Back for 'x'

We're not quite done yet. We've solved the integral in terms of 'u', but we need to express our answer in terms of the original variable, 'x'. Remember that we defined u = -3x. So, let's substitute -3x back in for 'u':

-2e^(u) + C becomes -2e^(-3x) + C

And there you have it! We've found the indefinite integral of 6e^(-3x).

Final Answer

The integral of 6e^(-3x) dx is:

∫ 6e^(-3x) dx = -2e^(-3x) + C

Where 'C' is the constant of integration.

Checking Our Work

It's always a good idea to check your work, especially in calculus. To do this, we can differentiate our answer and see if we get back the original function. Let's differentiate -2e^(-3x) + C with respect to x:

d/dx [-2e^(-3x) + C] = -2 * d/dx [e^(-3x)] + d/dx [C]

The derivative of a constant 'C' is 0, so the second term disappears.

To differentiate e^(-3x), we use the chain rule:

d/dx [e^(-3x)] = e^(-3x) * d/dx [-3x] = e^(-3x) * -3 = -3e^(-3x)

Substituting this back in, we get:

-2 * (-3e^(-3x)) = 6e^(-3x)

And that's exactly our original function! So, we know we've integrated correctly.

Key Takeaways

Let's recap the key steps we took to solve this integral:

  1. Simplified the integral: We pulled the constant 6 outside the integral.
  2. Used u-substitution: We identified u = -3x, found du, and substituted into the integral.
  3. Integrated the simplified expression: We integrated -2 ∫ e^(u) du to get -2e^(u) + C.
  4. Substituted back for 'x': We replaced 'u' with -3x to get our final answer: -2e^(-3x) + C.
  5. Checked our work: We differentiated our answer to verify that it matches the original function.

Practice Makes Perfect

Integration can seem tricky at first, but with practice, you'll get the hang of it. Try working through similar integrals using u-substitution. The more you practice, the easier it will become to identify the right substitutions and apply the integration rules. Remember, calculus is like learning a new language – it takes time and effort to become fluent.

Common Integration Techniques

Besides u-substitution, there are several other techniques you'll encounter in calculus, such as:

  • Integration by parts: Useful for integrals involving products of functions.
  • Trigonometric substitution: Used for integrals containing square roots of quadratic expressions.
  • Partial fraction decomposition: A technique for integrating rational functions.

Each of these methods has its own set of rules and applications, so it's important to become familiar with them. As you progress in your calculus journey, you'll learn when and how to apply each technique effectively.

Tips for Mastering Integration

Here are a few tips to help you master integration:

  • Understand the fundamentals: Make sure you have a solid grasp of differentiation rules before tackling integration.
  • Practice regularly: The more you practice, the better you'll become at recognizing patterns and applying the appropriate techniques.
  • Work through examples: Study solved examples to see how different integration techniques are applied.
  • Use resources: There are many excellent textbooks, websites, and online courses that can help you learn integration.
  • Don't be afraid to ask for help: If you're stuck, don't hesitate to ask your teacher, classmates, or online forums for assistance.

Conclusion

Integrating ∫ 6e^(-3x) dx involves a few key steps, including simplifying the integral, using u-substitution, and remembering the constant of integration. By breaking down the problem and understanding the underlying concepts, you can confidently tackle similar integrals. So, keep practicing, and you'll become an integration pro in no time! Keep up the great work, guys, and happy integrating!