Solving The Integral Of 3xe^(-x) A Step-by-Step Guide

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Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of integration to tackle a specific problem: evaluating the integral of 3xe^(-x) dx. This might seem daunting at first, but fear not! We'll break it down step by step, making sure you grasp every concept along the way. Whether you're a student grappling with calculus or simply a curious mind eager to expand your mathematical horizons, this guide is for you. So, let's roll up our sleeves and get started!

Understanding the Problem: Integration by Parts

At its core, this problem calls for a technique known as integration by parts. Now, why integration by parts, you might ask? Well, take a good look at our integrand, 3xe^(-x). We have a product of two functions here: a polynomial function (3x) and an exponential function (e^(-x)). When we encounter integrals involving products of different types of functions, integration by parts often comes to the rescue. This technique is essentially the reverse of the product rule for differentiation. Remember that golden oldie? The product rule states that the derivative of u times v is u'v + uv'. Integration by parts cleverly leverages this relationship to help us solve integrals that would otherwise be quite tricky. The formula for integration by parts is given by:

∫udv=uv−∫vdu\int u dv = uv - \int v du

The key to success with integration by parts lies in choosing the right u and dv. This choice can significantly impact the complexity of the problem. A wise selection can lead to a simpler integral, while a less strategic one might lead you down a rabbit hole of further integration. So, how do we make this crucial decision? A handy acronym, LIATE, often comes to the rescue. LIATE stands for:

  • Logarithmic functions
  • Inverse trigonometric functions
  • Algebraic functions
  • Trigonometric functions
  • Exponential functions

The order of LIATE suggests the priority for choosing u. In our case, 3x is an algebraic function and e^(-x) is an exponential function. According to LIATE, we should choose u as the algebraic function, 3x, and dv as the exponential function, e^(-x) dx. This choice is strategic because the derivative of 3x is simply 3, which is a constant. This will simplify the integral on the right-hand side of the integration by parts formula. On the other hand, if we had chosen u as e^(-x), its derivative would still be an exponential function, and we wouldn't have made much progress. Remember, the goal is to make the new integral, $\int v du$, easier to solve than the original one. So, with our u and dv selected, let's move on to the next step: applying the integration by parts formula.

Step-by-Step Solution: Applying Integration by Parts

Alright, guys, let's put our strategy into action and solve this integral step by step. We've already established that we'll use integration by parts, and we've made our crucial choices for u and dv. Let's recap those choices:

  • u = 3x
  • dv = e^(-x) dx

Now, we need to find du and v. Finding du is straightforward – it's simply the derivative of u with respect to x:

  • du = 3 dx

To find v, we need to integrate dv:

  • v = \int e^(-x) dx = -e^(-x)

Don't forget that little negative sign! It's a common pitfall to overlook, and it can throw off your entire calculation. Now that we have u, dv, du, and v, we're ready to plug them into the integration by parts formula:

∫udv=uv−∫vdu\int u dv = uv - \int v du

Substituting our values, we get:

∫3xe(−x)dx=(3x)(−e(−x))−∫(−e(−x))(3dx)\int 3x e^(-x) dx = (3x)(-e^(-x)) - \int (-e^(-x))(3 dx)

Let's simplify this a bit:

∫3xe(−x)dx=−3xe(−x)+∫3e(−x)dx\int 3x e^(-x) dx = -3xe^(-x) + \int 3e^(-x) dx

Notice how the integral on the right-hand side, $\int 3e^(-x) dx$, is much simpler than our original integral? This is the magic of integration by parts! We've effectively transformed a tricky integral into a more manageable one. Now, we just need to evaluate this new integral. Integrating 3e^(-x) with respect to x is relatively straightforward. We can pull the constant 3 out of the integral and integrate e^(-x), which we already know is -e^(-x):

∫3e(−x)dx=3∫e(−x)dx=3(−e(−x))=−3e(−x)\int 3e^(-x) dx = 3 \int e^(-x) dx = 3(-e^(-x)) = -3e^(-x)

Now, let's plug this back into our equation:

∫3xe(−x)dx=−3xe(−x)−3e(−x)+C\int 3x e^(-x) dx = -3xe^(-x) - 3e^(-x) + C

And there you have it! We've successfully integrated 3xe^(-x) dx. Notice the + C at the end? This is the constant of integration, a crucial reminder that the antiderivative of a function is not unique. We always need to add C to account for the possibility of a constant term that would disappear when we differentiate. So, our final answer is:

∫3xe(−x)dx=−3xe(−x)−3e(−x)+C\int 3x e^(-x) dx = -3xe^(-x) - 3e^(-x) + C

Simplifying the Result: Factoring and Presentation

While we've arrived at the correct answer, it's always a good practice to simplify our results as much as possible. In this case, we can factor out a common factor of -3e^(-x) from the two terms:

∫3xe(−x)dx=−3e(−x)(x+1)+C\int 3x e^(-x) dx = -3e^(-x)(x + 1) + C

This factored form is often considered a more elegant and concise way to present the solution. It highlights the underlying structure of the result and can be easier to work with in further calculations. Furthermore, simplifying our results can sometimes reveal hidden patterns or relationships that might not be immediately apparent in the unsimplified form. Think of it as polishing a gem to reveal its full brilliance! Beyond factoring, consider the overall presentation of your answer. Is it clear and easy to read? Are the mathematical symbols correctly formatted? A well-presented solution not only demonstrates your understanding of the math but also makes it easier for others (and your future self!) to follow your work. Remember, math is not just about finding the right answer; it's also about communicating your reasoning clearly and effectively. So, take pride in presenting your solutions in a neat and organized manner. It's a skill that will serve you well in all your mathematical endeavors. Now, let's recap the key takeaways from this journey of integration.

Key Takeaways and Practice Problems

Alright, guys, let's recap the key takeaways from our adventure in integrating 3xe^(-x) dx. We've covered quite a bit, from recognizing when to use integration by parts to the step-by-step application of the formula and finally, simplifying our result. Here's a quick rundown of the essential points:

  • Integration by parts is a powerful technique for integrating products of functions, especially when those functions are of different types (like algebraic and exponential, as in our example).
  • The formula for integration by parts is: $\int u dv = uv - \int v du$.
  • Choosing the right u and dv is crucial. The LIATE acronym (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) is a helpful guide for prioritizing your choices.
  • Remember to include the constant of integration, C, in your final answer. It's a small but vital detail!
  • Simplifying your result is always a good practice. Factoring out common factors can often lead to a more elegant and concise solution.

Now that we've solidified our understanding of the concepts, let's put our skills to the test with some practice problems. Practice is the key to mastering any mathematical technique, and integration by parts is no exception. The more you practice, the more comfortable and confident you'll become with the process. Here are a few problems to get you started:

  1. ∫xsin(x)dx\int x sin(x) dx
  2. ∫x2exdx\int x^2 e^x dx
  3. ∫ln(x)dx\int ln(x) dx
  4. ∫xcos(2x)dx\int x cos(2x) dx

Try tackling these problems using the steps we've outlined in this guide. Remember to carefully choose your u and dv, apply the integration by parts formula, and simplify your results. Don't be afraid to make mistakes – they're a natural part of the learning process. The important thing is to learn from your mistakes and keep practicing. Working through these problems will not only reinforce your understanding of integration by parts but also help you develop your problem-solving skills in general. And hey, if you get stuck, don't hesitate to revisit this guide or seek out other resources. There's a wealth of information available online and in textbooks to help you on your mathematical journey. Remember, the world of calculus is vast and fascinating, and every problem you solve is a step further on your path to mastery. So, keep practicing, keep exploring, and keep enjoying the beauty of mathematics! Until next time, happy integrating, guys!