Hey there, math explorers! Ever stumbled upon something that looks a little… intimidating? Like, a whole bunch of symbols doing a fancy dance, and you're just standing there, scratching your head? Well, today we're going to peek at something like that, but don't worry, we're going to take it slow and make it as chill as a Sunday morning. We're talking about evaluating an integral. Specifically, this one: the integral of 3 times the arctangent of 1/x, from x = 1 to x = 9. Yeah, I know, it sounds like a secret code for mathematicians, right? But stick with me, because there's some really neat stuff happening under the hood!
So, what's this "integral" thing, anyway? Think of it like this: if differentiation is about finding the slope of a curve at any given point, integration is like figuring out the area under that curve. Imagine you've got a wiggly line on a graph – integration helps us find the space it encloses with the x-axis. Pretty cool, huh? It's like finding the volume of an oddly shaped swimming pool or the total distance a car has traveled based on its speed over time. It's a way to add up infinitely many tiny pieces to get a whole picture.
And what's this "arctangent" business? You know how sine, cosine, and tangent are functions that relate angles in a right triangle to its side lengths? Arctangent, often written as arctan or tan⁻¹, is the inverse of the tangent function. So, if you tell arctangent a number, it tells you what angle would have that tangent value. Think of it as asking, "Okay, if the slope of this line is, say, 2, what's the angle it makes with the horizontal?" It’s like reverse-engineering an angle from a ratio.
Now, let's look at our specific problem: the integral of 3 * arctan(1/x) dx from 1 to 9. That "3" out front? Easy peasy. It's just a constant multiplier. When we integrate, we can just pull constants out to the front and deal with them later. So, we’re essentially looking at 3 times the area under the curve of arctan(1/x) between our starting point (x=1) and our ending point (x=9).
The real star of the show here is arctan(1/x). Why is this particular function interesting? Well, let's play around with it a little. What happens when x is really, really big? Like, if x is 1000, then 1/x is 0.001. The arctangent of a tiny positive number is just a tiny positive angle, right? So, as x gets huge, arctan(1/x) gets closer and closer to zero. It's like a shy guest at a party, slowly fading into the background as the crowd grows.
Now, what about when x is really small, but positive? Like x = 0.001? Then 1/x is 1000! The arctangent of a really big number approaches π/2 (which is about 1.57). This is because as the tangent value gets infinitely large, the angle gets infinitely close to 90 degrees, or π/2 radians. So, as x approaches zero from the positive side, arctan(1/x) approaches π/2. It's like a really energetic guest who shouts louder and louder as they get closer to the microphone!
The function arctan(1/x) has this interesting behavior where it goes from near π/2 for small positive x to near 0 for large positive x. It's not a simple straight line or a gentle curve; it has a bit of a dramatic personality!
Okay, so how do we actually find this area? For many integrals, we have handy-dandy formulas. But for functions involving arctangent, especially with that 1/x inside, it's not always straightforward. We often need a technique called integration by parts. Don't let the name scare you! It's basically a clever way to rewrite an integral in a form that's easier to solve. It's like having a secret handshake that transforms a tricky puzzle into a simpler one.
Integration by parts comes from the product rule for differentiation. Remember the product rule? It helps us find the derivative of two functions multiplied together. Integration by parts is its cousin, helping us integrate a product of functions. The formula looks like this: ∫ u dv = uv - ∫ v du. We pick parts of our integrand to be 'u' and 'dv', differentiate 'u' to get 'du', and integrate 'dv' to get 'v'. It's a bit of a choose-your-own-adventure for integrals!
In our case, we have 3 * arctan(1/x). We can think of this as 3 * arctan(1/x) * 1. So, we need to decide what to call 'u' and what to call 'dv'. Often, a good strategy is to pick 'u' as the part that becomes simpler when differentiated, and 'dv' as the part that's easy to integrate. For arctan functions, it's usually a good idea to let 'u' be the arctan part itself, because its derivative is pretty standard.
So, let's set u = arctan(1/x). What's dv? Well, we need something to integrate. If we consider the whole expression, we could try setting dv = dx. Then the integral of dv is just v = x. Now we need to find du, the derivative of arctan(1/x). This requires the chain rule. The derivative of arctan(z) is 1/(1+z²). So, with z = 1/x, the derivative of arctan(1/x) is 1/(1 + (1/x)²) * (derivative of 1/x). The derivative of 1/x (or x⁻¹) is -x⁻² = -1/x². So, du = (1 / (1 + 1/x²)) * (-1/x²) dx. This looks a bit messy, doesn't it?
Let's simplify that expression for du. 1 + 1/x² can be written as (x² + 1)/x². So, 1 / (1 + 1/x²) is x² / (x² + 1). Multiplying this by -1/x², we get: du = (x² / (x² + 1)) * (-1/x²) dx = -1 / (x² + 1) dx. See? It simplified nicely!
Now, let's plug everything into our integration by parts formula: ∫ u dv = uv - ∫ v du.
We have: ∫ arctan(1/x) dx = arctan(1/x) * x - ∫ x * (-1 / (x² + 1)) dx.
This simplifies to: ∫ arctan(1/x) dx = x * arctan(1/x) + ∫ x / (x² + 1) dx.
We've transformed our original integral into something involving x * arctan(1/x) and a new integral, ∫ x / (x² + 1) dx. The new integral, ∫ x / (x² + 1) dx, looks much friendlier. This one we can solve with a simple substitution. Let w = x² + 1. Then dw = 2x dx, which means x dx = dw/2. So, the integral becomes ∫ (dw/2) / w = (1/2) ∫ 1/w dw = (1/2) ln|w| = (1/2) ln(x² + 1) (since x² + 1 is always positive).
Putting it all back together, the indefinite integral of arctan(1/x) dx is: x * arctan(1/x) + (1/2) ln(x² + 1) + C (don't forget our constant of integration, C, for indefinite integrals!).
But we're not done yet! We need to evaluate the definite integral from 1 to 9, and remember that original "3" we set aside? We need to multiply our result by 3.
So, we need to calculate: 3 * [x * arctan(1/x) + (1/2) ln(x² + 1)] evaluated from x=1 to x=9.
This means we plug in 9, then plug in 1, and subtract the second result from the first, and then multiply the whole thing by 3.
Let's break it down:
At x = 9:
9 * arctan(1/9) + (1/2) ln(9² + 1) = 9 * arctan(1/9) + (1/2) ln(82)
At x = 1:
1 * arctan(1/1) + (1/2) ln(1² + 1) = 1 * arctan(1) + (1/2) ln(2) = π/4 + (1/2) ln(2)
(Because arctan(1) is the angle whose tangent is 1, which is 45 degrees, or π/4 radians.)
Now, subtract the second from the first:
[9 * arctan(1/9) + (1/2) ln(82)] - [π/4 + (1/2) ln(2)]
And finally, multiply by 3:
3 * [9 * arctan(1/9) + (1/2) ln(82) - π/4 - (1/2) ln(2)]
This is the exact value of our integral! It might look like a mouthful, but it's a concrete number representing the precise area. It's like solving a complex recipe and ending up with a perfectly baked cake.
Why is this kind of thing cool? Because it shows us that even seemingly complicated mathematical expressions can be broken down, understood, and solved using clever techniques. It's a testament to the power of logic and a bit of creative problem-solving. It's like cracking a code, revealing a hidden pattern, or discovering a new pathway. And the fact that we can find an exact numerical value for the area under that arctan(1/x) curve between 1 and 9? That's pretty darn satisfying, don't you think?
So, the next time you see a bunch of symbols that look like a puzzle, remember that with a little curiosity and the right tools, you can often unravel their mysteries. Math isn't just about memorizing formulas; it's about exploring, discovering, and understanding the beautiful, logical world around us. Keep exploring, and who knows what amazing integrals you'll encounter next!
