Problem 73 - Calculus: Early Transcendentals

Question

Express the limit as a definite integral. \( \displaystyle \lim_{n \to \infty} \sum^{n}_{i = 1} \frac{i^4}{n^5} \) [\( Hint: \) Consider \( f(x) = x^4 \).]

Step-by-Step Solution

Answer

The limit expresses the definite integral \( \int_0^1 x^4 \, dx = \frac{1}{5} \).

Step 1: Identify the Function and Interval

The exercise provides the sum in the form of Riemann sums. We are told to consider the function \( f(x) = x^4 \). This tells us that we will be dealing with the function \( f(x) \) over an interval \([0, 1]\). In general, for limits representing definite integrals, the variable transformation involves \( f(x) \), and the division by \( n^5 \) indicates a \((b-a)/n\) term.

Step 2: Set Up the Riemann Sum

The given summation \( \sum_{i=1}^{n} \frac{i^4}{n^5} \) resembles the Riemann sum \( \sum_{i=1}^{n} f(x_i) \cdot \Delta x \) where \( x_i = \frac{i}{n} \) and \( \Delta x = \frac{1}{n} \). Substitute \( x_i = \frac{i}{n} \) into \( f(x) = x^4 \), giving \( f(x_i) = \left( \frac{i}{n} \right)^4 \).

Step 3: Perform the Substitution

Substitute \( f(\frac{i}{n}) \cdot \Delta x \) into the sum: \( \sum_{i=1}^{n} \left( \frac{i}{n} \right)^4 \cdot \frac{1}{n} = \sum_{i=1}^{n} \frac{i^4}{n^5} \). This confirms our earlier setup and allows us to equate this Riemann sum with the expression for the sum given in the problem.

Step 4: Express as a Definite Integral

Recognize that the expression \( \sum_{i=1}^{n} \left( \frac{i}{n} \right)^4 \cdot \frac{1}{n} \) represents the Riemann sum for the definite integral of \( f(x) = x^4 \) over \( x = 0 \) to \( x = 1 \) as \( n \to \infty \). Therefore, \[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{i^4}{n^5} = \int_0^1 x^4 \, dx. \]

Step 5: Evaluate the Definite Integral

To complete the task of expressing the limit as a definite integral, we may evaluate it. Compute \( \int_0^1 x^4 \, dx \). The antiderivative of \( x^4 \) is \( \frac{x^5}{5} \). Thus, evaluate \( \left[ \frac{x^5}{5} \right]_0^1 = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5} \).

Key Concepts

Riemann sum

The Riemann sum is a powerful concept used to approximate the area under a curve, which is integral to understanding definite integrals. In essence, it involves summing up slices of a curve, over a defined interval, to reach an approximation of the total area. This method involves:

• Selecting a function, such as our given case, where \( f(x) = x^4 \).
• Dividing the area of interest, typically an interval like \([0, 1]\), into \( n \) sub-intervals or partitions.
• Calculating the height of each rectangle using \( f(x_i) \), where \( x_i = \frac{i}{n} \), a point in our sub-interval.
• Determining the width of each rectangle, noted as \( \Delta x = \frac{1}{n} \).

Through this approach, each slice or rectangle has a tiny width due to \( \Delta x \), and as \( n \) approaches infinity, the sum of these small areas, \( \sum_{i=1}^{n} f(x_i) \cdot \Delta x \), converges to the exact area under the curve, which is our definite integral.

limit of a function

The limit of a function plays a crucial role when working with Riemann sums and definite integrals. It describes what happens to a function as the input approaches a particular point. In our study of Riemann sums, this specifically refers to what occurs as \( n \) approaches infinity. Here's how it is relevant in our context:

• As \( n \to \infty \), the partitions of our interval become infinitesimally small.
• The Riemann sum converges to a single value, representing the area under the curve of \( f(x) = x^4 \).
• This limit is essentially the definite integral, \( \int_0^1 x^4 \, dx \).

Understanding limits allows us to transition from a sum of approximations to the precise total area under a function over an interval.

integration

Integration is the inverse process of differentiation and is used to find areas under curves, among other things. In our case, it allows us to calculate the exact area under the function \( f(x) = x^4 \) from 0 to 1. Performing this integration involves:

• Finding the antiderivative of the function, which here is \( \frac{x^5}{5} \).
• Substituting the bounds of the interval, producing \( \left[ \frac{x^5}{5} \right]_0^1 \).
• Evaluating the resulting expression to find the area, hence \( \frac{1}{5} \).

Through integration, we elegantly move from the sum approximation with Riemann sums to the exact solution of a definite integral. This provides a concrete value, representing the accumulated area under a given function over a set interval.

x^4 function

The function \( f(x) = x^4 \) is a simple yet insightful function, characterized by an increasing curvature. It serves as a great example for calculating integrals and applying the concept of Riemann sums. Let's break it down:

• The function's formula, \( x^4 \), indicates rapid growth, as small changes in \( x \) result in larger changes to \( x^4 \).
• On the interval \([0, 1]\), \( x^4 \) starts at 0 and moves smoothly to 1.
• Its integration from 0 to 1 involves finding its antiderivative, which is \( \frac{x^5}{5} \), and computing the definite integral becomes fairly straightforward.

Understanding this function and its properties lays the groundwork for interpreting how and why the Riemann sums approach an exact value when \( n \to \infty \). It's a classic choice for exercises in introductory calculus due to its relative simplicity and the essential calculation principles it embodies.