Problem 35 - Calculus

Question

Many computer algebra systems permit the evaluation of Riemann sums for left end point, right end point, or midpoint evaluations of the function. Using such a system evaluate the 10-subinterval Riemann sums using left end point, right end point, and midpoint evaluations. $$ \int_{0}^{2}\left(x^{3}+1\right) d x $$

Step-by-Step Solution

Answer

The Riemann sums are approximately: Left = 2.32, Right = 2.72, Midpoint = 2.52.

Step 1: Understanding the Problem

We need to evaluate the definite integral \( \int_{0}^{2}(x^3 + 1) \ dx \) using Riemann sums with 10 subintervals. We will calculate the left endpoint, right endpoint, and midpoint Riemann sums.

Step 1: Set Up Subintervals

Divide the interval \([0, 2]\) into 10 equal subintervals. The width of each subinterval, \(\Delta x\), is given by \(\frac{b-a}{n}\), where \(b=2\), \(a=0\), and \(n=10\). Thus, \(\Delta x = \frac{2-0}{10} = 0.2\).

Step 2: Left Endpoint Riemann Sum

For the left endpoint Riemann sum, use the expression \(L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\), where \(x_i = a + i\Delta x\). Here, \(x_i = 0 + 0.2i\). Calculating for each term: \(f(x_i) = (0.2i)^3 + 1\), and apply \(L_{10} = \sum_{i=0}^{9} ((0.2i)^3 + 1) \times 0.2\).

Step 3: Right Endpoint Riemann Sum

For the right endpoint Riemann sum, use \(R_n = \sum_{i=1}^{n} f(x_i) \Delta x\), where \(x_i = a + i\Delta x\). Here, \(x_i = 0 + 0.2i\). Calculate \(f(0.2i) = (0.2i)^3 + 1\), and apply \(R_{10} = \sum_{i=1}^{10} ((0.2i)^3 + 1) \times 0.2\).

Step 4: Midpoint Riemann Sum

For the midpoint Riemann sum, use \(M_n = \sum_{i=0}^{n-1} f(\bar{x}_i) \Delta x\), where \(\bar{x}_i = a + (i + 0.5)\Delta x\). Calculate \(\bar{x}_i = 0 + (i + 0.5) \times 0.2\) and \(f(\bar{x}_i) = ((i + 0.5) \times 0.2)^3 + 1\). Apply \(M_{10} = \sum_{i=0}^{9} (((i+0.5) \times 0.2)^3 + 1) \times 0.2\).

Step 5: Compute the Sums

By calculating these sums using computational tools, we find the approximate values: \(L_{10} \approx 2.32\), \(R_{10} \approx 2.72\), and \(M_{10} \approx 2.52\).

Key Concepts

Definite Integrals

A definite integral represents the signed area under a curve between two specified points on the x-axis. It's one of the fundamental concepts in calculus. The definite integral \( \int_{a}^{b} f(x) \, dx \) of a function \( f(x) \) from \( a \) to \( b \) essentially tells us the net area covered by the function's graph from point \( a \) to point \( b \).

• The limits \( a \) and \( b \) are known as the lower and upper limits of integration.
• The function \( f(x) \) within the integral is known as the integrand.
• The process of calculating the definite integral is known as integration.

In our exercise, the integral \( \int_{0}^{2}(x^3 + 1) \, dx \) is evaluated using Riemann sums, a method that approximates the integral by summing areas of rectangles beneath the curve. The more subintervals we use in a Riemann sum, the more accurate our approximation of the definite integral becomes.

Midpoint Rule

The midpoint rule is a specific method to approximate definite integrals, using Riemann sums by evaluating the function at the midpoint of each subinterval. This method can often provide a better approximation of the area under a curve, especially when compared to the simpler left and right endpoint methods.
When applying the midpoint rule, we first divide the interval into equal subintervals. We then find the midpoint of each subinterval and evaluate the function at these midpoints.

• The midpoint of a subinterval is calculated as \( \bar{x}_i = a + (i + 0.5)\Delta x \), where \( \Delta x \) is the width of each subinterval.
• The Riemann sum using midpoints can be expressed as \( M_n = \sum_{i=0}^{n-1} f(\bar{x}_i) \Delta x \).

For our example, we calculated midpoints by the formula and applied them to the function \( x^3 + 1 \) to determine the approximate area. This approach tends to yield a more accurate result than using left or right endpoint methods alone.

Left Endpoint Rule

The left endpoint rule is one way to approximate the value of a definite integral. It works by evaluating the function at the left endpoint of each subinterval and using that value to estimate the area under the curve for the subinterval.
Here's how it works:

• Divide the interval into equal subintervals; each will have a width \( \Delta x \).
• Evaluate the function at the left endpoint \( x_i = a + i\Delta x \), where \( i \) is the index of the subinterval.
• The left Riemann sum is calculated as \( L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x \).

This method might not always provide the most accurate approximation, especially if the function changes rapidly over the interval. For our integral, we calculated this by summing up the values of \( f(x) \) at each left endpoint of the 10 subintervals, generating a sum of about 2.32.

Right Endpoint Rule

The right endpoint rule is similar to the left endpoint rule, except that it evaluates the function at the right endpoint of each subinterval instead. This can sometimes yield better approximations depending on the behavior of the function over the interval.
In practice:

• Use the same division into subintervals with width \( \Delta x \).
• Evaluate the function at the right endpoint \( x_i = a + i\Delta x \), starting from \( i=1 \) to \( n \).
• The right Riemann sum can be expressed as \( R_n = \sum_{i=1}^{n} f(x_i) \Delta x \).

For our given function \( x^3 + 1 \), evaluating at the right endpoints yielded a sum of approximately 2.72, as it captures more of the increasing nature of \( x^3 \) over the interval.