Problem 75 - Calculus: Early Transcendentals

Question

Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on \( [0, 1] \). \( \displaystyle \lim_{n \to \infty} \sum_{i = 1}^n \biggl( \frac{i^4}{n^5} + \frac{i}{n^2} \biggr) \)

Step-by-Step Solution

Answer

The limit of the sum is \( \frac{7}{10} \).

Step 1: Understanding the Riemann Sum

First, recognize that the given sum can be rewritten in terms of a Riemann sum. The expression is \( \lim_{n \to \infty} \sum_{i=1}^n \left( \frac{i^4}{n^5} + \frac{i}{n^2} \right) \). The general form of a Riemann sum for a function \( f(x) \) on the interval \([0,1]\) is \( \lim_{n \to \infty} \sum_{i=1}^n f\left(\frac{i}{n}\right) \cdot \frac{1}{n} \). Here, \( \Delta x = \frac{1}{n} \) and \( x_i = \frac{i}{n} \).

Step 2: Express the Sum as a Single Function

Separate the sum into two individual sums and express them inside the limit as functions of \(\frac{i}{n}\):\[\lim_{n \to \infty} \sum_{i=1}^n \left( \left( \frac{i}{n} \right)^4 \cdot \frac{1}{n} \right) + \lim_{n \to \infty} \sum_{i=1}^n \left( \frac{i}{n} \cdot \frac{1}{n} \right) \]This represents two distinct functions: \( f(x) = x^4 \) and \( g(x) = x \) over the interval \([0, 1]\).

Step 3: Evaluate the First Integral

The first term \( \lim_{n \to \infty} \sum_{i=1}^n \left( \left( \frac{i}{n} \right)^4 \cdot \frac{1}{n} \right) \) is equivalent to the integral \( \int_0^1 x^4 \, dx \). Evaluating the integral:\[\int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5} - 0 = \frac{1}{5}\]

Step 4: Evaluate the Second Integral

The second term \( \lim_{n \to \infty} \sum_{i=1}^n \left( \frac{i}{n} \cdot \frac{1}{n} \right) \) is equivalent to the integral \( \int_0^1 x \, dx \). Evaluating the integral gives:\[\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} - 0 = \frac{1}{2}\]

Step 5: Add the Results of the Integrals

Combine the results from Step 3 and Step 4 to find the total:\[\frac{1}{5} + \frac{1}{2} = \frac{2}{10} + \frac{5}{10} = \frac{7}{10}\]

Step 6: Final Conclusion

Therefore, the limit of the sum is the sum of both integrals, which equals \( \frac{7}{10} \). Thus, the original expression simplifies to \( \frac{7}{10} \).

Key Concepts

Definite Integral

A definite integral represents the accumulation of quantities over a specified interval. For example, in the given function, we have two parts that need evaluation: \( f(x) = x^4 \) and \( g(x) = x \), both over the interval from 0 to 1. The definite integral here gives us a precise numerical value, showing the total area under the function's curve.

The notation \( \int_a^b f(x) \, dx \) tells us to integrate the function \( f(x) \) starting at \( x = a \) and ending at \( x = b \). For instance:

• In the case of \( f(x) = x^4 \), we calculate \( \int_0^1 x^4 \, dx \) to find the exact area under the curve from 0 to 1.
• Similarly, for \( g(x) = x \), we calculate \( \int_0^1 x \, dx \).

The definite integral sums these infinitesimally small areas to form the complete area under the curve.

Limit Evaluation

Limits help us understand the behavior of a function as it approaches a certain point or infinity. The Riemann sum involves a limit, as it considers infinitely many partitions of the interval.

In our example, we are evaluating the limit:

• \( \lim_{n \to \infty} \sum_{i=1}^n \left( \frac{i^4}{n^5} + \frac{i}{n^2} \right) \).

This means we divide the interval [0,1] into \( n \) subintervals, taking the limit as \( n \) approaches infinity. This process enables us to convert the sum into a definite integral.

Limits transform the discrete sum into a continuous process, allowing the computation of exact values even though the process seems infinitely complex.

Integral Calculus

Integral Calculus is the branch of mathematics focused on accumulation and area calculation. It provides the tools to evaluate Riemann sums and find the value of definite integrals.

Using the principles of integral calculus, we can:

• Convert a sum expression, such as the Riemann sum, into an integral.
• Evaluate the integral using antiderivatives and boundary conditions.

This is demonstrated with two functions under consideration:

• Find \( \int_0^1 x^4 \, dx = \frac{1}{5} \)
• Find \( \int_0^1 x \, dx = \frac{1}{2} \)

With integral calculus, we derive these precise results from potentially complex sums, summarizing an infinite number of elements in a simple form.

Function Analysis

Function Analysis involves studying functions to understand their behavior, growth, and areas under their curves. For this exercise, analyzing each function, \( x^4 \) and \( x \), helps simplify the complex Riemann sum into manageable pieces.

Here are a few key steps:

• Recognize the forms of the individual functions within the given sum.
• Understand their role over the interval [0,1].
• Simplify your Riemann sum by separating and evaluating each function's contribution individually.

Function analysis allows such partitioning and evaluation to effectively integrate and sum the individual components for a complete understanding.