Evaluating ∫(x-8)ln(x) Dx And Verification By Differentiation

Hey guys! Today, we're diving into a fun calculus problem: evaluating the integral of (x-8)ln(x) and then double-checking our answer by differentiating it. It's a classic technique to ensure we've nailed the integration. So, grab your pencils, and let's get started! Remember that we are assuming that x > 0.

The Integral Problem: ∫(x-8)ln(x) dx

So, the main goal is to find the indefinite integral of the function (x-8)ln(x). This means we're looking for a function whose derivative is (x-8)ln(x). Now, when we look at this integral, it's pretty clear that we can't just apply a simple power rule or a basic integration formula. We've got a product of two different types of functions here: a polynomial (x-8) and a logarithmic function (ln(x)). This is a big hint that we should be using integration by parts. Integration by parts is a technique derived from the product rule for differentiation, and it's super handy when you're dealing with integrals involving products of functions.

The formula for integration by parts is:

∫u dv = uv - ∫v du

The trick here is to choose our u and dv wisely. The goal is to pick a u that simplifies when we differentiate it and a dv that's easy to integrate. In our case, ln(x) is a great candidate for u because its derivative is 1/x, which is much simpler. That means our dv will be (x-8)dx. Let's break it down:

• Let u = ln(x)
• Then du = (1/x) dx
• Let dv = (x-8) dx
• Then v = ∫(x-8) dx = (1/2)x^2 - 8x

Now we have all the pieces we need to plug into our integration by parts formula. Let's do it!

∫(x-8)ln(x) dx = ln(x) * [(1/2)x^2 - 8x] - ∫[(1/2)x^2 - 8x] * (1/x) dx

Okay, that looks a bit intimidating, but don't worry, we'll simplify it step by step. First, let's clean up the expression a bit:

∫(x-8)ln(x) dx = [(1/2)x^2 - 8x]ln(x) - ∫[(1/2)x - 8] dx

See? It's already looking better. Now we have a much simpler integral to deal with: ∫[(1/2)x - 8] dx. This is something we can easily integrate using the power rule.

∫[(1/2)x - 8] dx = (1/4)x^2 - 8x + C

Where 'C' is the constant of integration, which we always need to include for indefinite integrals. Now we can substitute this back into our main equation:

∫(x-8)ln(x) dx = [(1/2)x^2 - 8x]ln(x) - [(1/4)x^2 - 8x] + C

And there we have it! We've found the integral. To summarize, the integral of (x-8)ln(x) dx is [(1/2)x^2 - 8x]ln(x) - [(1/4)x^2 - 8x] + C. But we're not done yet! The next step is super important: we need to verify our answer. And how do we do that? By differentiating, of course!

Verifying the Result by Differentiation

Alright, guys, we've got our integral result, and now it's time to put on our detective hats and verify if our hard work paid off. The best way to check an indefinite integral is by differentiating the result. If we've done our integration correctly, the derivative of our result should be the original function we started with, which is (x-8)ln(x). So, let's dive into differentiating this expression:

Our result from the integration was:

F(x) = [(1/2)x^2 - 8x]ln(x) - [(1/4)x^2 - 8x] + C

We need to find F'(x), the derivative of F(x). This involves using a few different differentiation rules, including the product rule and the power rule. Remember that the product rule states that if we have a function that's a product of two other functions, say u(x)v(x), then its derivative is:

(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)

And the power rule is a classic: if we have x^n, its derivative is nx^(n-1). Let's start by differentiating the first part of our function, [(1/2)x^2 - 8x]ln(x). We'll use the product rule here:

• Let u(x) = (1/2)x^2 - 8x
• Then u'(x) = x - 8
• Let v(x) = ln(x)
• Then v'(x) = 1/x

Now, applying the product rule:

([(1/2)x^2 - 8x]ln(x))' = (x - 8)ln(x) + (1/2)x^2 - 8x

Let's simplify that a bit:

= (x - 8)ln(x) + (1/2)x - 8

Great! We've differentiated the first part. Now let's differentiate the second part, which is [(1/4)x^2 - 8x]. This is a straightforward application of the power rule:

([(1/4)x^2 - 8x])' = (1/2)x - 8

And the derivative of our constant of integration, 'C', is simply 0, since constants don't change.

Now we can put it all together. The derivative of our entire function F(x) is:

F'(x) = (x - 8)ln(x) + (1/2)x - 8 - [(1/2)x - 8] + 0

Notice that we're subtracting the derivative of the second part because it was subtracted in our original integrated function. Let's simplify this expression further:

F'(x) = (x - 8)ln(x) + (1/2)x - 8 - (1/2)x + 8

Hey, look at that! The (1/2)x terms cancel out, and the -8 and +8 also cancel out. We're left with:

F'(x) = (x - 8)ln(x)

And that, my friends, is exactly the original function we started with! This means our integration was spot on. We've successfully evaluated the integral and verified our result by differentiation. High five!

Key Takeaways and Tips for Integration by Parts

Alright, so we've tackled a pretty neat integral problem today, and I hope you're feeling more confident about integration by parts. But before we wrap up, let's quickly recap some key takeaways and tips that will help you nail these kinds of problems in the future.

1. Choosing u and dv Wisely: The heart of integration by parts lies in selecting the right u and dv. Remember, the goal is to pick a u that simplifies when you differentiate it and a dv that's easy to integrate. A handy acronym to remember is LIATE, which stands for:

   • Logarithmic functions (like ln(x), log(x))
   • Inverse trigonometric functions (like arctan(x), arcsin(x))
   • Algebraic functions (polynomials like x, x^2, x-8)
   • Trigonometric functions (like sin(x), cos(x))
   • Exponential functions (like e^x, 2^x)

   This order gives you a general guideline for choosing u. Functions higher on the list are usually good candidates for u because their derivatives tend to be simpler.

2. Don't Be Afraid to Integrate by Parts Multiple Times: Sometimes, you might need to apply integration by parts more than once to solve a single integral. If the new integral you get after the first application still involves a product of functions, it's a sign you might need to use the technique again. Keep at it, and you'll get there!

3. Simplification is Your Friend: After each step in integration by parts, take a moment to simplify the resulting expression. This can make the next steps much easier and reduce the chance of making mistakes. Combine like terms, factor out common factors, and generally tidy things up.

4. Verification is Crucial: As we demonstrated today, verifying your result by differentiation is an essential step. It's like having a built-in error check. If the derivative of your answer doesn't match the original integrand, you know something went wrong, and you can go back and review your work.

5. Practice Makes Perfect: Like any mathematical technique, the more you practice integration by parts, the better you'll become at it. Work through a variety of examples, and you'll start to develop an intuition for which functions are good choices for u and dv.

Common Pitfalls to Avoid

Let's quickly touch on some common mistakes people make when using integration by parts so you can steer clear of them:

• Incorrectly Applying the Formula: Double-check that you're plugging everything into the integration by parts formula correctly. It's easy to mix up the u, dv, du, and v terms.
• Forgetting the Constant of Integration: Always, always, always add the constant of integration 'C' when you're evaluating indefinite integrals. It's a small detail, but it's super important.
• Choosing the Wrong u and dv: This is the most common pitfall. If your choice of u and dv leads to a more complicated integral, try swapping them around or using a different approach.
• Making Algebraic Errors: Be careful with your algebra and arithmetic. It's easy to make small mistakes when simplifying expressions, especially when dealing with fractions and negative signs.

So, there you have it! Integration by parts can seem tricky at first, but with a solid understanding of the formula, careful selection of u and dv, and plenty of practice, you'll be mastering these integrals in no time. Keep up the great work, and happy integrating, guys!

In summary, evaluating integrals like ∫(x-8)ln(x) dx requires a strategic approach, and integration by parts is a powerful tool for such cases. Remember the importance of choosing appropriate parts for u and dv, simplifying expressions, and verifying your results through differentiation. By mastering these techniques and avoiding common pitfalls, you'll enhance your calculus skills and confidently tackle complex integration problems.