Problem 83 - Biocalculus Calculus for the Life Sciences

Question

\(83-84\) Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on \([0,1]\) . \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{i^{3}}{n^{4}}\)

Step-by-Step Solution

Answer

The limit is \( \frac{1}{4} \).

Step 1: Recognize the Riemann Sum

We start by identifying that the given expression is in the form of a Riemann sum. The expression \( \sum_{i=1}^{n} \frac{i^{3}}{n^{4}} \) can be rewritten as \( \sum_{i=1}^{n} \frac{1}{n} \times \left( \frac{i}{n} \right)^{3} \). This is the form \( \sum_{i=1}^{n} f\left( \frac{i}{n} \right) \frac{1}{n} \), where \( f(x) = x^{3} \). This is the Riemann sum for the function \( f(x) = x^3 \) on the interval \([0, 1]\).

Step 2: Determine the Integral

Since we recognized that the sum represents the Riemann integral of \( f(x) = x^3 \) over \([0, 1]\), we write the corresponding definite integral: \( \int_{0}^{1} x^{3} \, dx \).

Step 3: Integrate the Function

Now, integrate \( f(x) = x^3 \) over the interval \([0, 1]\). The integral is \( \int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} \).

Step 4: Evaluate the Integral

Evaluating the integral \( \left[ \frac{x^4}{4} \right]_{0}^{1} \) gives us \( \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \).

Step 5: Conclude the Limit

Therefore, the limit of the Riemann sum as \( n \) approaches infinity is the same as the value of the integral. So \( \lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{i^{3}}{n^{4}} = \frac{1}{4} \).

Key Concepts

Definite Integral

A definite integral is a mathematical concept used to find the total accumulation of quantities. It represents the area under the curve of a function over a given interval. In the context of the original exercise, we deal with the expression for calculating the area under the curve \( f(x) = x^3 \) on the interval \([0, 1]\).

• The definite integral provides a way to quantify the total sum of infinitely small changes or slices over this interval.
• Mathematically, it is denoted by \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, depicting the range over which you evaluate the function.

In our exercise, the definite integral \( \int_{0}^{1} x^3 \, dx \) translates to computing the sum of the function's values over its domain. This results in a weighted area, considering the entire interval's length. By solving this, we found that the integral equals \( \frac{1}{4} \). This reveals the cumulative total under \( f(x) = x^3 \) from zero to one.

Limit Evaluation

Limit evaluation is a crucial process in understanding the behavior of sequences or functions as a parameter approaches a certain value. In the given exercise, we examined the limit as \( n \) tends to infinity. This limit helps us determine what the sum \( \sum_{i=1}^{n} \frac{i^{3}}{n^{4}} \) approaches when the number of subdivisions become very large.

• To evaluate a limit, especially in cases like this, we use techniques such as transforming sums into integrals or using L'Hopital's Rule when appropriate.
• In our context, transforming the sum into a recognizable Riemann sum facilitates seeing it as a definite integral problem, aligning with the conceptual grasp of calculus.

The transforming process gives us \( \lim_{n \to \infty} \sum_{i=1}^{n} \frac{i^{3}}{n^{4}} = \int_{0}^{1} x^3 \, dx \). Evaluating this limit is essential to transition from finite approximations to exact, continuous evaluations indicative of calculus applications.

Riemann Integral

The Riemann integral is a fundamental concept in calculus, providing a formal method to define the integral of a function over an interval. This method involves taking the limit of a sum of areas of rectangles as the number of subdivisions increases, and the width of subdivisions decreases.

• The process starts with dividing the interval \([a, b]\) into \( n \) equal parts, making a partition, and then crafting rectangles under the curve based on the height determined by the function, \( f(x) \).
• The Riemann sum \( \sum_{i=1}^{n} f\left( \frac{i}{n} \right) \frac{1}{n} \) associates the function values with these partition points, accumulating approximate areas.
• As \( n \to \infty \), the sum approaches a perfect representation of the region under the curve \( f(x) \).

In our task, recognizing the sum as a Riemann integral allows us to understand \( \sum_{i=1}^{n} \frac{i^{3}}{n^{4}} \) as a way to approximate \( \int_{0}^{1} x^3 \, dx \). Essentially, this enlightenment helps us transition from approximate computations to exact analysis.