Expressing Limits As Definite Integrals: A Cotangent Example

Hey guys! Today, we're diving into a super cool concept in calculus: expressing a limit of a Riemann sum as a definite integral. This is like translating from one mathematical language to another, and it's seriously powerful. We'll be tackling a specific example involving the cotangent function, so buckle up and let's get started!

Understanding the Problem

Our mission, should we choose to accept it (and we do!), is to take the following limit and rewrite it as a definite integral:

$$\lim _{\|P\| \rightarrow 0} \sum_{k=1}^n(\cot c_k) \Delta x_k$$

Where:

• $\|P\|$ represents the norm (or mesh) of the partition P, which essentially means the length of the longest subinterval in the partition.
• The limit $\|P\| \rightarrow 0$ tells us that we're making the subintervals in our partition infinitely small.
• The summation $\sum_{k=1}^n$ is a sum from k = 1 to n, where n is the number of subintervals in our partition.
• $\cot c_k$ is the cotangent function evaluated at a point $c_k$ within the k-th subinterval.
• $\Delta x_k$ represents the width of the k-th subinterval.
• We are given that P is a partition of the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$.

Before we jump into the solution, let's break down why this is important and what it really means. This expression is in the form of a Riemann sum. Riemann sums are fundamental to the definition of the definite integral. They represent an approximation of the area under a curve by dividing the area into rectangles and summing their areas. The limit as the width of these rectangles approaches zero gives us the exact area, which is the definite integral.

So, the core idea here is that this limit is a definite integral in disguise! Our job is to unmask it and write it in its integral form. This involves recognizing the function being evaluated, the interval of integration, and setting up the definite integral notation correctly. Understanding this connection between Riemann sums and definite integrals is crucial for grasping the fundamental theorem of calculus, which provides a powerful way to evaluate definite integrals using antiderivatives.

The Riemann Sum and the Definite Integral

Let’s dive deeper into the connection between Riemann sums and definite integrals. This is where the magic happens, guys! The definite integral is formally defined as the limit of a Riemann sum. This might sound a bit abstract, so let’s break it down. Imagine you have a curve, and you want to find the area under that curve between two points on the x-axis. One way to approximate this area is to divide the region under the curve into a bunch of rectangles.

Think of it like tiling a floor, but instead of perfect square tiles, you’re using rectangles that may not perfectly fit under the curve. The sum of the areas of these rectangles is a Riemann sum. Now, if you make these rectangles infinitely thin (that's what the limit as the norm of the partition goes to zero means), the approximation becomes exact. This limit of the sum is the definite integral.

Mathematically, it looks like this:

$$\lim_{\|P\| \rightarrow 0} \sum_{k=1}^n f(c_k) \Delta x_k = \int_a^b f(x) dx$$

Where:

• f(x) is the function we're integrating (the curve we're finding the area under).
• [a, b] is the interval of integration (the section of the x-axis we're interested in).
• $c_k$ is a point in the k-th subinterval.
• $\Delta x_k$ is the width of the k-th subinterval.

This formula is the key to solving our problem. We need to match the given limit to the left side of this equation and then identify the corresponding components on the right side to write the definite integral. It's like having a decoder ring – we just need to figure out which part of the Riemann sum corresponds to each element of the definite integral.

In our case, we have $\lim _{\|P\| \rightarrow 0} \sum_{k=1}^n(\cot c_k) \Delta x_k$. Can you see how this lines up with the general form of a Riemann sum? The function f(x) is hiding in plain sight, and the interval of integration is given. Let's move on to identifying these key components in the next section!

Identifying the Function and Interval

Okay, let’s put on our detective hats and figure out the pieces of the puzzle! Remember our general form of the limit of a Riemann sum:

$$\lim_{\|P\| \rightarrow 0} \sum_{k=1}^n f(c_k) \Delta x_k$$

And our specific limit:

$$\lim _{\|P\| \rightarrow 0} \sum_{k=1}^n(\cot c_k) \Delta x_k$$

It's like a mathematical version of "spot the difference," isn't it? By comparing these two expressions, we can clearly see that the function f(x) in our case is the cotangent function, or $\cot x$. This is the function we'll be integrating. So, f(x) = cot(x).

Now, what about the interval of integration? The problem statement gives us a crucial clue: "P is a partition of $[\frac{\pi}{6}, \frac{\pi}{3}]$". This tells us that the interval over which we're summing these rectangles, and therefore the interval of integration for our definite integral, is $[\frac{\pi}{6}, \frac{\pi}{3}]$. So, a = π/6 and b = π/3.

We've successfully identified the two key ingredients for our definite integral: the function and the interval! This is a huge step forward. It's like finding the coordinates to a hidden treasure – we're getting closer to unearthing the definite integral.

To recap, we have:

• f(x) = cot(x)
• a = π/6
• b = π/3

With this information in hand, we're ready to write the limit as a definite integral. Let’s do it!

Expressing the Limit as a Definite Integral

Alright, guys, the moment we've been waiting for! We've identified the function and the interval, and now we're ready to write our limit as a definite integral. This is where everything comes together and we see the beautiful connection between Riemann sums and integrals.

Remember the general form:

$$\lim_{\|P\| \rightarrow 0} \sum_{k=1}^n f(c_k) \Delta x_k = \int_a^b f(x) dx$$

We know:

• f(x) = cot(x)
• a = π/6
• b = π/3

So, we simply substitute these values into the definite integral notation:

$$\int_a^b f(x) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cot(x) dx$$

And there you have it! We've successfully expressed the limit of the Riemann sum as a definite integral:

$$\lim _{\|P\| \rightarrow 0} \sum_{k=1}^n(\cot c_k) \Delta x_k = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cot(x) dx$$

Isn't that neat? We've taken a somewhat intimidating limit of a sum and transformed it into a compact and elegant integral. This is the power of the definite integral – it provides a concise way to represent the accumulation of infinitely small quantities.

Now, while we've expressed the limit as a definite integral, we haven't actually evaluated the integral. That would be the next step if we wanted to find the numerical value of the area under the cotangent curve between π/6 and π/3. Evaluating this integral would involve finding the antiderivative of cot(x), which we can tackle using a little trigonometric trickery and substitution. But for now, our mission is accomplished – we've successfully expressed the limit as a definite integral!

Conclusion

So, there you have it, guys! We've taken a journey from a limit of a Riemann sum to a definite integral, showcasing a fundamental concept in calculus. We started by understanding the problem, then we dove into the connection between Riemann sums and definite integrals, identified the function and interval, and finally, expressed the limit as a definite integral.

This process highlights the power of the definite integral as a tool for representing limits of sums. It's like having a mathematical Swiss Army knife – it can help us solve a wide range of problems involving accumulation and continuous change.

Remember, the key takeaway here is that the definite integral is the limit of a Riemann sum. By recognizing this connection, we can translate between these two representations and unlock a deeper understanding of calculus. Keep practicing these types of problems, and you'll become a pro at spotting those hidden integrals! You've got this!