Indefinite Integral: $\int X\sqrt{x+6}dx$ Solved

Hey Plastik Magazine readers! Let's dive into the fascinating world of calculus, specifically tackling indefinite integrals. Today, we're going to break down how to solve the integral of $x \sqrt{x+6} dx$. Don't worry, even if you're not a math whiz, I'll walk you through it step-by-step. We'll find the indefinite integral and then, for a little extra fun, we'll check our work by differentiating. Sounds good? Awesome, let's get started!

The Quest for $\int x\sqrt{x+6}dx$: A Step-by-Step Journey

Okay, guys, the first thing we're gonna do is choose a u-substitution. This is a super handy trick for simplifying integrals. The goal is to pick something that makes the rest of the integral easier to deal with. In this case, a great choice is: $u = x + 6$. Why? Because the square root is messing everything up. Let's start with $u = x + 6$, then we need to find $du$. Differentiating both sides with respect to x, we get $du = dx$. Easy peasy! Now, we also need to express x in terms of u. From $u = x + 6$, we can rearrange to get $x = u - 6$. Perfect, we're all set to make our substitution. Replacing everything in the original integral, we now have $\int (u - 6) \sqrt{u} du$. See? Much cleaner already. This simplifies to $\int (u^{3/2} - 6u^{1/2}) du$. This is a lot more manageable. This is because it transforms the integral into something we can readily solve using the power rule for integration. Now we integrate term by term. For the first term, $\int u^{3/2} du$, we add 1 to the exponent (3/2 + 1 = 5/2) and divide by the new exponent: $\frac{u^{5/2}}{5/2} = \frac{2}{5}u^{5/2}$. For the second term, $\int 6u^{1/2} du$, we do the same thing: add 1 to the exponent (1/2 + 1 = 3/2) and divide by the new exponent: $\frac{6u^{3/2}}{3/2} = 4u^{3/2}$. Don't forget the constant of integration, C, because this is an indefinite integral. Putting it all together, we have $\frac{2}{5}u^{5/2} - 4u^{3/2} + C$. But wait! Remember that our original integral was in terms of x. So, we need to substitute back x for u. Since $u = x + 6$, we replace u everywhere in our answer. This gives us $\frac{2}{5}(x + 6)^{5/2} - 4(x + 6)^{3/2} + C$. And that, my friends, is our indefinite integral! We've successfully solved $\int x\sqrt{x+6}dx$. Before we move on to verify the result, take a moment to appreciate the journey! This process of strategically transforming and simplifying the integral is fundamental in calculus.

Now, before we move on to the next part, let's take a quick breather and just appreciate what we've done. We've taken a seemingly complex integral and, through smart substitution and the power rule, have found a solution. Pretty cool, huh? But we're not quite done yet... We still need to check our work!

Verifying the Integral: Differentiating for Accuracy

Alright, time to check our answer! To make sure we're correct, we're going to differentiate our result, $\frac{2}{5}(x + 6)^{5/2} - 4(x + 6)^{3/2} + C$, and see if it gets us back to the original integrand, $x\sqrt{x+6}$. Ready? Let's go! We'll differentiate each term separately using the chain rule. Remember, the chain rule is your best friend here. For the first term, $\frac{2}{5}(x + 6)^{5/2}$, we bring down the exponent (5/2), multiply it by the coefficient, and then reduce the exponent by 1 (5/2 - 1 = 3/2). So we get: $\frac{2}{5} * \frac{5}{2}(x + 6)^{3/2} = (x + 6)^{3/2}$. The derivative of the constant term, C, is simply 0. Remember to multiply the derivative of the inside function (x+6)' which is 1. Now, for the second term, $-4(x + 6)^{3/2}$, we bring down the exponent (3/2), multiply it by the coefficient, and reduce the exponent by 1 (3/2 - 1 = 1/2). This gives us $-4 * \frac{3}{2}(x + 6)^{1/2} = -6(x + 6)^{1/2}$. So, our derivative is $(x + 6)^{3/2} - 6(x + 6)^{1/2}$. But we're not done yet! We need to simplify this expression to see if it matches our original integrand, $x\sqrt{x+6}$. Let's factor out the $\sqrt{x + 6}$ term: $(x + 6)\sqrt{x+6} - 6\sqrt{x+6} = \sqrt{x+6} [(x + 6) - 6] = \sqrt{x+6} * x = x\sqrt{x+6}$. And there you have it! Our derivative does indeed match the original integrand. This confirms that our indefinite integral is correct. This is a very satisfying moment. It's a reminder that all the steps we took were valid, that all the calculations were correct, and that we have successfully solved the problem. We have not only found the integral but also validated our answer through the process of differentiation. This process is a testament to the interconnectedness of calculus concepts and it reinforces the value of checking your work.

Key Takeaways and Final Thoughts

So, what have we learned, guys? We've successfully found the indefinite integral of $x\sqrt{x+6}dx$ using u-substitution and the power rule for integration. We then checked our work by differentiating our result and verifying that it matched the original integrand. This process highlights the relationship between integration and differentiation and it reinforces the importance of verification in problem-solving. Remember, practice makes perfect. The more integrals you solve, the better you'll get. Don't be afraid to try different techniques and to check your answers. And remember, calculus can be a lot of fun once you get the hang of it. This whole process of solving, verifying, and celebrating your success is what makes math so enjoyable. Now that you've got this under your belt, try some similar problems. You can change the function or adjust the bounds. The important part is that you grasp the concepts and feel comfortable applying them. Keep experimenting, and keep challenging yourself! Also, consider that in many real-world applications, this concept is utilized in optimization problems where you have to find the maximum or minimum value of a function. The ability to find integrals is critical for determining how quantities change over time or space. So go out there, solve some integrals, and have fun! The world of calculus is open to you. Feel free to explore related topics such as definite integrals, integration by parts, and trigonometric substitutions. Keep learning, keep exploring, and keep the mathematical spirit alive! You guys are awesome. Keep up the good work, and thanks for joining me on this math adventure.