Limit Of Summed Areas A Calculus Problem Solution

Hey there, math enthusiasts! Today, we're diving deep into a fascinating calculus problem that involves calculating the limit of a sum of areas. This is not just your run-of-the-mill integral question; it elegantly combines integration, series, and limits, making it a truly rewarding challenge. So, buckle up, and let’s embark on this mathematical journey together!

The Problem at Hand

We are given a sequence of areas, denoted by Sₙ, where each Sₙ represents the area of the region bounded by the curve y = x²(1 - x)ⁿ, the x-axis, and the interval 0 ≤ x ≤ 1. Our mission, should we choose to accept it, is to find the limit of the sum of these areas as n approaches infinity. Mathematically, this is expressed as:

$$\lim_{n \to \infty} \sum_{k=1}^{n} S_k = \frac{1}{\sqrt{A}}$$

Our ultimate goal is to determine the value of A. Sounds intriguing, right? Let’s break it down step by step.

Step 1 Calculating Sₙ

The first logical step is to find a general expression for Sₙ. Remember, the area under a curve can be found using integration. Therefore, Sₙ can be calculated as:

$$S_n = \int_{0}^{1} x^2(1 - x)^n dx$$

This integral might look a bit intimidating at first, but don't worry; we have tools to tackle it! The most effective technique here is integration by parts. For those who need a refresher, integration by parts stems from the product rule for differentiation and is expressed as:

$$\int u dv = uv - \int v du$$

The key is to choose u and dv strategically to simplify the integral. In our case, a wise choice is:

• u = x²
• dv = (1 - x)ⁿ dx

Differentiating u and integrating dv, we get:

• du = 2x dx
• v = -\frac{(1 - x)^{n+1}}{n + 1}

Now, let’s apply the integration by parts formula:

$$S_n = \left[-x^2\frac{(1 - x)^{n+1}}{n + 1}\right]_0^1 + \int_{0}^{1} \frac{2x(1 - x)^{n+1}}{n + 1} dx$$

The first term elegantly vanishes at both limits (x = 0 and x = 1), simplifying our expression to:

$$S_n = \frac{2}{n + 1} \int_{0}^{1} x(1 - x)^{n+1} dx$$

We're not done yet! We need to apply integration by parts again to the new integral. This time, we choose:

• u = x
• dv = (1 - x)^{n+1} dx

Differentiating and integrating, we have:

• du = dx
• v = -\frac{(1 - x)^{n+2}}{n + 2}

Applying integration by parts once more:

$$\int_{0}^{1} x(1 - x)^{n+1} dx = \left[-x\frac{(1 - x)^{n+2}}{n + 2}\right]_0^1 + \int_{0}^{1} \frac{(1 - x)^{n+2}}{n + 2} dx$$

Again, the first term vanishes at the limits, leaving us with:

$$\int_{0}^{1} x(1 - x)^{n+1} dx = \frac{1}{n + 2} \int_{0}^{1} (1 - x)^{n+2} dx$$

The remaining integral is straightforward:

$$\int_{0}^{1} (1 - x)^{n+2} dx = \left[-\frac{(1 - x)^{n+3}}{n + 3}\right]_0^1 = \frac{1}{n + 3}$$

Plugging this back into our expression for Sₙ, we get:

$$S_n = \frac{2}{n + 1} \cdot \frac{1}{n + 2} \cdot \frac{1}{n + 3} = \frac{2}{(n + 1)(n + 2)(n + 3)}$$

Finally, we have a manageable expression for Sₙ! It’s a rational function, and we're one step closer to solving the puzzle.

Step 2 Summing the Series

Now that we have Sₙ, we need to find the sum of the series:

$$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} \frac{2}{(k + 1)(k + 2)(k + 3)}$$

This series might seem daunting, but there's a clever trick we can use: partial fraction decomposition. This technique allows us to break down a complex rational function into simpler fractions that are easier to sum.

We want to express Sₖ in the form:

$$\frac{2}{(k + 1)(k + 2)(k + 3)} = \frac{A}{k + 1} + \frac{B}{k + 2} + \frac{C}{k + 3}$$

To find A, B, and C, we multiply both sides by (k + 1)(k + 2)(k + 3):

$$2 = A(k + 2)(k + 3) + B(k + 1)(k + 3) + C(k + 1)(k + 2)$$

We can solve for A, B, and C by strategically choosing values of k:

• Let k = -1: 2 = A(1)(2) => A = 1
• Let k = -2: 2 = B(-1)(1) => B = -2
• Let k = -3: 2 = C(-2)(-1) => C = 1

So, we can rewrite Sₖ as:

$$S_k = \frac{1}{k + 1} - \frac{2}{k + 2} + \frac{1}{k + 3}$$

Now, let’s plug this back into our summation:

$$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} \left(\frac{1}{k + 1} - \frac{2}{k + 2} + \frac{1}{k + 3}\right)$$

This looks much more manageable! We can now expand the sum and observe a beautiful phenomenon: telescoping. Telescoping series are series where terms cancel out in a cascading manner, simplifying the sum.

Let’s write out the first few terms and the last few terms to see the cancellation pattern:

$$\begin{aligned} \sum_{k=1}^{n} S_k &= \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right) \\ &+ \left(\frac{1}{3} - \frac{2}{4} + \frac{1}{5}\right) \\ &+ \left(\frac{1}{4} - \frac{2}{5} + \frac{1}{6}\right) \\ &+ \cdots \\ &+ \left(\frac{1}{n} - \frac{2}{n + 1} + \frac{1}{n + 2}\right) \\ &+ \left(\frac{1}{n + 1} - \frac{2}{n + 2} + \frac{1}{n + 3}\right) \end{aligned}$$

Notice how many terms cancel out? The -2/3 cancels with 1/3 + 1/3 from the neighboring terms. 1/4 cancels with other terms. After the dust settles, we are left with:

$$\sum_{k=1}^{n} S_k = \frac{1}{2} - \frac{1}{3} - \frac{1}{n + 2} + \frac{1}{n + 3}$$

Simplifying this, we get:

$$\sum_{k=1}^{n} S_k = \frac{1}{6} - \frac{1}{n + 2} + \frac{1}{n + 3}$$

Step 3 Taking the Limit

We're in the home stretch! Now we need to find the limit as n approaches infinity:

$$\lim_{n \to \infty} \sum_{k=1}^{n} S_k = \lim_{n \to \infty} \left(\frac{1}{6} - \frac{1}{n + 2} + \frac{1}{n + 3}\right)$$

As n approaches infinity, the terms 1/(n + 2) and 1/(n + 3) both approach 0. Therefore,

$$\lim_{n \to \infty} \sum_{k=1}^{n} S_k = \frac{1}{6}$$

Step 4 Solving for A

We were given that:

$$\lim_{n \to \infty} \sum_{k=1}^{n} S_k = \frac{1}{\sqrt{A}}$$

We found that the limit is 1/6, so:

$$\frac{1}{6} = \frac{1}{\sqrt{A}}$$

Squaring both sides and inverting, we get:

$$\sqrt{A} = 6$$

$$\frac{1}{\sqrt{A}} = \frac{1}{6}$$

Squaring both sides of $\sqrt{A} = 6$, we finally obtain:

$$A = 36$$

Conclusion

Eureka! We have successfully navigated the intricate world of calculus, integration by parts, series, and limits to find that A = 36. This problem showcases the beauty and power of calculus in solving complex problems. Remember, the key is to break down the problem into smaller, manageable steps, apply the appropriate techniques, and persevere through the calculations. Whether you are studying integration or diving deep into sequence and series, understanding these concepts will significantly level up your calculus game.

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Calculus Problem Solving Limit of Summed Areas Solution

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How to solve the limit of the sum of areas problem where Sn is the area between y=(x^2)(1-x)n; 0≤x≤1 and the x-axis, and lim (n→∞) Σ(k=1 to n) Sk = 1/√A. Find A.