Evaluate The Integral. 3 5 Arctan 1 X Dx 1

Ever stared at a mathematical expression that looks like a secret code and wondered what all the fuss is about? Well, sometimes those codes unlock incredible insights and powerful tools! Today, we're going to dive into a particularly satisfying mathematical puzzle: evaluating the integral of 3 * arctan(1/x) with respect to x, from 1 to 5. Think of it as a treasure hunt where the treasure is a precise numerical value, and the map is the elegant world of calculus. This isn't just about abstract numbers; understanding how to tackle integrals like this is fundamental to so many real-world applications, from predicting the path of a projectile to designing efficient circuits and understanding complex economic models. It’s a way of summing up continuous change, and that’s a pretty big deal!

Why This Integral is a Fun Challenge

You might be thinking, "Integrals? Fun?" Absolutely! For those who enjoy a good puzzle, integrating a function that involves an inverse trigonometric function like arctangent presents a delightful challenge. It’s like a sophisticated lock-picking exercise for your brain. The function itself, arctan(1/x), has some interesting behavior. As 'x' gets very large, 1/x gets very small, and arctan(1/x) approaches 0. As 'x' gets smaller (approaching 0), 1/x gets very large, and arctan(1/x) approaches π/2. This behavior hints at the rich landscape of curves and areas that calculus helps us understand. The '3' in front of it simply scales the area we're interested in, making it three times as tall, but the core challenge lies in the arctan(1/x).

The Purpose: Unveiling the Area Under the Curve

At its heart, evaluating an integral is about finding the area bounded by a function's curve, the x-axis, and specific vertical lines. In our case, we're looking for the precise area under the curve of the function f(x) = 3 * arctan(1/x) between the vertical lines x=1 and x=5. This is called a definite integral because we have specific limits of integration. It tells us the total accumulation of the function's value over that interval. Imagine filling that space with an infinitesimally fine liquid; the integral tells you the exact volume of that liquid!

The Benefits: Powering Predictions and Designs

The benefits of mastering integral calculus, and solving problems like this one, are immense. For engineers, it's crucial for calculating work done, fluid pressure, and the displacement of objects. Physicists use it to derive laws of motion and understand complex phenomena like electromagnetism. Economists employ it to model cumulative growth and predict market trends. Even in computer graphics, integrals play a role in rendering realistic images. By solving this specific integral, we’re practicing a technique that is a building block for solving far more complex, real-world problems. It’s about developing the analytical skills to break down complicated situations into manageable parts and then reconstruct them to find meaningful answers.

Our Mathematical Journey Begins

So, how do we actually get to that numerical treasure? The standard approach for integrals involving products of functions, especially those with inverse trigonometric functions, is integration by parts. This is a powerful technique derived from the product rule of differentiation. The formula for integration by parts is: ∫ u dv = uv - ∫ v du. The art lies in choosing which part of our integrand to assign to 'u' and which to 'dv'. A common heuristic is the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), which suggests the order in which to pick 'u'. In our case, arctan(1/x) falls into the 'Inverse trigonometric' category, making it a prime candidate for 'u'.

Let's set:

u = arctan(1/x)

dv = 3 dx

From here, we need to find 'du' (the derivative of u) and 'v' (the integral of dv). Differentiating arctan(1/x) requires a bit of chain rule, as we're differentiating arctan of a function of x. The derivative of arctan(y) is 1/(1+y²). So, the derivative of arctan(1/x) will involve 1/(1 + (1/x)²) multiplied by the derivative of (1/x), which is -1/x². This leads to:

du = (1 / (1 + (1/x)²)) * (-1/x²) dx

du = (x² / (x² + 1)) * (-1/x²) dx

du = -1 / (x² + 1) dx

And integrating dv = 3 dx is straightforward:

v = ∫ 3 dx = 3x

Now, we plug these back into our integration by parts formula: ∫ u dv = uv - ∫ v du.

∫ 3 arctan(1/x) dx = arctan(1/x) * (3x) - ∫ (3x) * (-1 / (x² + 1)) dx

∫ 3 arctan(1/x) dx = 3x arctan(1/x) + ∫ (3x / (x² + 1)) dx

Tackling the Remaining Integral

We're left with a new integral: ∫ (3x / (x² + 1)) dx. This one can be solved using a simple u-substitution. Let w = x² + 1. Then, dw = 2x dx. We can rewrite our integral in terms of 'w':

∫ (3x / (x² + 1)) dx = (3/2) ∫ (2x / (x² + 1)) dx

= (3/2) ∫ (1 / w) dw

= (3/2) ln|w| + C

= (3/2) ln(x² + 1) + C

Remembering to substitute back w = x² + 1. So, our indefinite integral for 3 arctan(1/x) is:

∫ 3 arctan(1/x) dx = 3x arctan(1/x) + (3/2) ln(x² + 1) + C

Evaluating the Definite Integral

Now for the grand finale: applying the limits of integration from 1 to 5. We need to evaluate our result at x=5 and subtract its value at x=1.

[3x arctan(1/x) + (3/2) ln(x² + 1)] from 1 to 5

At x=5:

3(5) arctan(1/5) + (3/2) ln(5² + 1) = 15 arctan(1/5) + (3/2) ln(26)

At x=1:

3(1) arctan(1/1) + (3/2) ln(1² + 1) = 3 arctan(1) + (3/2) ln(2)

Since arctan(1) = π/4, this becomes: 3(π/4) + (3/2) ln(2) = (3π/4) + (3/2) ln(2)

Subtracting the value at x=1 from the value at x=5:

[15 arctan(1/5) + (3/2) ln(26)] - [(3π/4) + (3/2) ln(2)]

= 15 arctan(1/5) - (3π/4) + (3/2) ln(26) - (3/2) ln(2)

Using logarithm properties, ln(26) - ln(2) = ln(26/2) = ln(13).

So the final result is: 15 arctan(1/5) - (3π/4) + (3/2) ln(13)

And there you have it! A precise numerical value representing the area under our curve. It’s a beautiful blend of algebraic manipulation and calculus principles, showcasing how we can conquer complex mathematical expressions to reveal concrete answers. This exercise, while specific, equips you with the foundational techniques to tackle a vast array of integration problems, making you a more powerful problem-solver in the process!