Evaluating ∫[1,2] Ln(s)/s^6 Ds Using Integration By Parts

Hey guys! Ever stumbled upon an integral that looks like it's written in another language? You stare at it, and it stares back, leaving you wondering where to even begin? Well, today, we're going to tackle one of those integrals head-on, turning it from a mathematical monster into a manageable problem. We're diving deep into the world of integration by parts, a super-powerful technique that's like the Swiss Army knife of calculus. Our mission, should we choose to accept it, is to evaluate the definite integral ∫[1,2] ln(s)/s^6 ds. Buckle up, because this is going to be a fun ride!

What is Integration by Parts? Your Calculus Superpower

Before we jump into the nitty-gritty of our specific integral, let's take a moment to understand what integration by parts is all about. Think of it as the reverse of the product rule for differentiation. Remember that old friend? The product rule tells us how to differentiate the product of two functions. Integration by parts does the opposite – it helps us integrate the product of two functions.

The formula for integration by parts is the key here:

∫u dv = uv - ∫v du

Now, this might look a bit cryptic at first, but let's break it down. The integral we're trying to solve is on the left side (∫u dv). The right side gives us a new way to express that integral. We've essentially transformed our original integral into something (uv) minus another integral (∫v du). The trick is to choose our 'u' and 'dv' wisely so that the new integral (∫v du) is easier to solve than the original one. That's the magic of integration by parts!

Choosing Your 'u' and 'dv' The Secret Sauce

The most crucial step in integration by parts is selecting which part of your integrand (the function you're integrating) will be 'u' and which part will be 'dv'. There's no one-size-fits-all rule here, but there's a helpful mnemonic that many calculus students swear by: LIATE.

LIATE stands for:

• Logarithmic functions (like ln(x))
• Inverse trigonometric functions (like arctan(x))
• Algebraic functions (like x^2, x^3, etc.)
• Trigonometric functions (like sin(x), cos(x))
• Exponential functions (like e^x)

The idea is that you should choose your 'u' based on this order. If you have a logarithmic function and an algebraic function, the logarithmic function is usually a good choice for 'u'. This is because the derivative of a logarithmic function is often simpler than the function itself, which is what we want when we calculate 'du'. On the other hand, 'dv' should be something that you can easily integrate. You see, integration by parts is a game of strategic simplification. You want to break down your integral into pieces that you can handle more easily.

Applying Integration by Parts to ∫[1,2] ln(s)/s^6 ds Let's Get to Work!

Okay, enough theory! Let's get our hands dirty with our specific integral: ∫[1,2] ln(s)/s^6 ds. Remember, our goal is to use integration by parts to evaluate this definite integral.

Step 1: Choosing 'u' and 'dv'

Following the LIATE rule, we have a logarithmic function (ln(s)) and an algebraic function (1/s^6). Logarithmic functions come before algebraic functions in LIATE, so we'll choose:

u = ln(s)

This means that the rest of the integrand must be 'dv':

dv = (1/s^6) ds = s^(-6) ds

Step 2: Finding 'du' and 'v'

Now we need to find 'du' and 'v'. 'du' is simply the derivative of 'u':

du = (1/s) ds

To find 'v', we need to integrate 'dv':

v = ∫s^(-6) ds = -1/5 * s^(-5)

Remember the power rule for integration? We add 1 to the exponent and divide by the new exponent. Don't forget the constant of integration when doing indefinite integrals, but in integration by parts, we typically omit it until the very end (it'll cancel out anyway!).

Step 3: Plugging into the Formula

Now comes the exciting part – plugging our 'u', 'v', 'du', and 'dv' into the integration by parts formula:

∫u dv = uv - ∫v du

So, for our integral:

∫[1,2] ln(s)/s^6 ds = [ln(s) * (-1/5 * s^(-5))][from 1 to 2] - ∫[1,2] (-1/5 * s^(-5)) * (1/s) ds

Let's simplify this a bit:

∫[1,2] ln(s)/s^6 ds = [-ln(s) / (5s^5)][from 1 to 2] + (1/5) ∫[1,2] s^(-6) ds

Step 4: Evaluating the New Integral

Look at that! The integral on the right-hand side is much simpler than our original one. We can easily integrate s^(-6):

(1/5) ∫[1,2] s^(-6) ds = (1/5) * [-1/5 * s^(-5)][from 1 to 2] = [-1 / (25s^5)][from 1 to 2]

Step 5: Putting It All Together and Evaluating the Definite Integral

Now let's plug this back into our equation and evaluate the definite integral:

∫[1,2] ln(s)/s^6 ds = [-ln(s) / (5s^5)][from 1 to 2] + [-1 / (25s^5)][from 1 to 2]

To evaluate a definite integral, we plug in the upper limit (2) and subtract the result of plugging in the lower limit (1):

= [-ln(2) / (5 * 2^5) - 1 / (25 * 2^5)] - [-ln(1) / (5 * 1^5) - 1 / (25 * 1^5)]

Remember that ln(1) = 0, so this simplifies to:

= [-ln(2) / 160 - 1 / 800] - [- 0 - 1/25]

= -ln(2) / 160 - 1 / 800 + 1/25

To get a common denominator, let's use 800:

= (-5ln(2) - 1 + 32) / 800

= (31 - 5ln(2)) / 800

And there you have it! The value of the definite integral ∫[1,2] ln(s)/s^6 ds is (31 - 5ln(2)) / 800. We conquered it using the mighty power of integration by parts!

Key Takeaways and Pro Tips for Mastering Integration by Parts

Wow, we've covered a lot! Let's recap the key steps and throw in a few extra tips to help you become an integration by parts master:

• Master the Formula: The formula ∫u dv = uv - ∫v du is your best friend. Memorize it, love it, live it.
• LIATE is Your Guide: Use the LIATE mnemonic to help you choose 'u'. Remember, it's just a guideline, not a strict rule. Sometimes you might need to experiment.
• Simpler is Better: The goal is to make the new integral (∫v du) easier to solve. If it's not, you might need to rethink your choice of 'u' and 'dv'.
• Don't Forget the Signs: Pay close attention to those negative signs! A small sign error can throw off your entire calculation.
• Practice Makes Perfect: The more you practice integration by parts, the better you'll get at recognizing when to use it and how to choose 'u' and 'dv' effectively.
• Definite Integrals Require Careful Evaluation: Remember to plug in the limits of integration correctly and handle the arithmetic carefully.

Beyond the Basics Real-World Applications and Advanced Techniques

We've seen how integration by parts can tackle integrals that seem impossible at first glance. But the power of this technique extends far beyond textbook examples. It's used extensively in various fields, including:

• Physics: Calculating work done by a variable force, finding the center of mass of an object.
• Engineering: Analyzing circuits, solving differential equations.
• Probability and Statistics: Calculating expected values and variances.
• Economics: Modeling economic growth and financial markets.

As you delve deeper into calculus, you'll encounter more advanced techniques that build upon integration by parts. One such technique is tabular integration, also known as the