Problem 2 - Essential Calculus

Question

In Exercises 1 and 2, use Example 1 as a model to evaluate the limit $$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(c_{i}\right) \Delta x_{i}$$ over the region bounded by the graphs of the equations. $$ \begin{array}{l} f(x)=2 \sqrt[3]{x}, \quad y=0, \quad x=0, \quad x=1 \\ \text { (Hint: Let } \left.c_{i}=i^{3} / n^{3} .\right) \end{array} $$

Step-by-Step Solution

Answer

The value of the limit is 3.

Step 1: Define The Variables

First of all, it is important to understand what each variable means. \(f(x)\) is the function to be integrated. \(x\) represents the x-values inside the region. Negotiable \(x\) values range between 0 and 1 which also defines the limits of integration. Here, the task states \(c_{i} = i^3/n^3\) which will be the average values of \(x\) in the interval we will consider while the term \(\Delta x_{i}\) represents the width of each interval.

Step 2: Apply the Definitions

Next, we apply the definitions of the given variables and perform substitutions into the given limit. As a result, we get: \[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f\left(\dfrac{i^3}{n^3}\right) \Delta x_{i} = \lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} f\left(\dfrac{i^3}{n^3}\right)\]Here, we assumed the width of small intervals, \(\Delta x_{i}\), approaches 1/n as \(n\) tends to infinity.

Step 3: Implement the Function

The function \(f(x)\) is defined as \(2\sqrt[3]{x}\), so replace \(f\left(\dfrac{i^3}{n^3}\right)\) with \(2\sqrt[3]{\dfrac{i^3}{n^3}}\):\[\lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} 2\sqrt[3]{\dfrac{i^3}{n^3}}\]

Step 4: Simplify the Expression

The expression inside the square root simplifies to \(\dfrac{i}{n}\):\[\lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} 2\sqrt[3]{\dfrac{i}{n}}\]

Step 5: Finding The Limit

Now, by summing over all \(i\), we multiply by \(2/n\) (since there are \(n\) terms in the sum) and take the limit as \(n\) approaches infinity. This converts the sum into an integral from 0 to 1 of \(2\sqrt[3]{x}\) dx. Thus, we get: \[\lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{i=1}^{n} 2\sqrt[3]{\dfrac{i}{n}} = \int_{0}^{1} 2\sqrt[3]{x}\, dx\]

Step 6: Calculating The Integral

Finally, calculate the integral, which gives the final result:\[\int_{0}^{1} 2\sqrt[3]{x}\, dx = \left[3x^{4/3}\right]_0^1 = 3 - 0 = 3\]

Key Concepts

Limit Evaluation

Limit evaluation is a fundamental aspect of calculus. It involves finding the value that a function approaches as the input approaches a particular value. In the context of integration using Riemann sums, the limit helps us to approximate the area under a curve by increasing the number of subintervals to infinity.

In this problem, the limit evaluation is critical to translating a Riemann sum into a definite integral. As the number of divisions, represented by \(n\), approaches infinity, each subinterval becomes infinitesimally small. This is expressed in our formula as \(\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f\big(c_{i}\big) \Delta x_{i}\).

• Increasing \(n\) refines the approximation of the area under the curve.
• As \(n\) tends to infinity, the approximation becomes exact, converting the sum into an integral.
• This is a core step in calculus for moving from discrete sums to continuous integrals.

Riemann Sum

The Riemann sum is a method of approximating integrals, which is essentially the area under a curve on a graph. It involves dividing the domain into small subintervals and summing up the product of function values at sample points and the widths of these intervals.

In our exercise, the sum \(\sum_{i=1}^{n} f\left(\frac{i^3}{n^3}\right) \Delta x_{i}\) represents the Riemann sum:

• Each term in the sum corresponds to an area element in the approximation.
• \(c_i = i^3/n^3\) is chosen to represent the sample point in each subinterval.
• The term \(\Delta x_{i}\) denotes the width of intervals, approaching \(1/n\) as \(n\) increases.

Understanding this concept is crucial for grasping how integrals are estimated using finite sums.

Definite Integral

A definite integral is an important concept in calculus used to calculate the exact area under a curve between two points on the x-axis. In this problem, the transition from a Riemann sum to a definite integral involves taking the limit of the sum expression.

Our aim is to compute \(\int_{0}^{1} 2\sqrt[3]{x}\, dx\), which represents the area under the function \(f(x) = 2\sqrt[3]{x}\) from \(x = 0\) to \(x = 1\):

• Representing an infinite sum as an integral simplifies computation and provides exact results.
• The limits of integration, 0 and 1 in this case, define the interval over which we seek the area.
• Understanding how to convert sums to integrals is key to applying calculus to real-world scenarios.

Integral Evaluation

Integral evaluation refers to the process of calculating the value of a definite integral, which gives the exact area under a curve within specified limits. In this exercise, the integration of \(2\sqrt[3]{x}\) from 0 to 1 insightfully demonstrates the culmination of translating a Riemann sum into a meaningful area.

The integration process entails finding the antiderivative of the function \(2\sqrt[3]{x}\) which is \(3x^{4/3}\), then evaluating this at the given limits:

• The antiderivative simplifies calculations: \([3x^{4/3}]_0^1 = 3(1) - 3(0) = 3\).
• This result confirms the area under the curve is exactly 3, as computed from the integral.
• Integral evaluation is vital for making sense of real-world phenomena described by functions.

By mastering this, students can efficiently connect theory with practical applications.