Problem 14 - Applied Calculus

Question

Use a graphing calculator Riemann Sum program from the Internet (see page 332 ) to find the following Riemann sums. i. Calculate the Riemann sum for each function for the following values of \(n: 10,100,\) and \(1000 .\) Use left, right, or midpoint rectangles, making a table of the answers, rounded to three decimal places. ii. Find the exact value of the area under the curve by evaluating an appropriate definite integral using the Fundamental Theorem. The values of the Riemann sums from part (i) should approach this number. $$ f(x)=x^{2}+1 \text { from } a=0 \text { to } b=1 $$

Step-by-Step Solution

Answer

The Riemann sums approach the exact integral value of \(\frac{4}{3}\) as \(n\) increases.

Step 1: Setup Riemann Sums

Begin by setting up the Riemann Sums for the given function \(f(x) = x^2 + 1\) over the interval \([0,1]\). Choose left, right, or midpoint rectangles for approximation. In this exercise, we will use the midpoint method for accuracy.

Step 2: Calculate Riemann Sum n=10

Using a graphing calculator program, divide the interval \([0,1]\) into 10 equal subintervals of width \(\Delta x = 0.1\). Calculate the midpoint of each subinterval and evaluate \(f(x)\) at each midpoint. Sum all these values and multiply by \(\Delta x\) to approximate the area.

Step 3: Calculate Riemann Sum n=100

Repeat the process for \(n=100\). This time, divide the interval into 100 subintervals, each with width \(\Delta x = 0.01\). Evaluate \(f(x)\) at the midpoints of these subintervals, sum them, and multiply by \(\Delta x\).

Step 4: Calculate Riemann Sum n=1000

Perform the same procedure for \(n=1000\), dividing the interval into 1000 subintervals of width \(\Delta x = 0.001\). Find the midpoint of each subinterval, evaluate \(f(x)\) at these points, sum them, and multiply by \(\Delta x\).

Step 5: Table of Riemann Sums

Compile the results of the Riemann sum calculations into a table:\[\begin{array}{|c|c|}\hlinen & \text{Riemann Sum} \\hline10 & \text{(Result from Step 2)} \100 & \text{(Result from Step 3)} \1000 & \text{(Result from Step 4)} \\hline\end{array}\]Ensure all values are rounded to three decimal places.

Step 6: Exact Value by Definite Integral

Determine the exact area under the curve using integration. Integrate \(f(x) = x^2 + 1\) from 0 to 1 using the Fundamental Theorem of Calculus:\[\int_{0}^{1} (x^2 + 1) \, dx = \left[ \frac{x^3}{3} + x \right]_0^1\]Evaluate the integral to find the exact area.

Step 7: Comparison of Results

Compare the Riemann sums with the exact integral value. As \(n\) increases, the Riemann sum values should approach the definite integral value, confirming accuracy.

Key Concepts

Definite Integral

A definite integral is a powerful mathematical concept used to compute the exact area under a curve between two specified limits on a graph. In the context of the exercise, we focus on the function \( f(x) = x^2 + 1 \) over the interval \([0, 1]\). The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration.
The purpose of evaluating a definite integral is to find the exact value of the area, contrary to an estimate. This exact value gives us important information about the function's behavior over the specified range. In this case, we are interested in precisely calculating:

• 
  \( \int_{0}^{1} (x^2 + 1) \, dx = \left[ \frac{x^3}{3} + x \right]_0^1 \).
• Substitute the limits to get \( \left( \frac{1^3}{3} + 1 \right) - \left( \frac{0^3}{3} + 0 \right) = \frac{4}{3} \).
• Thus, the exact area is \( \frac{4}{3} \) square units.

This definite integral result will serve as a benchmark to evaluate the accuracy of the Riemann sum approximations.

Midpoint Approximation

Midpoint approximation is a technique used in the calculation of Riemann sums, which are approximations of definite integrals. When using midpoint approximation, we divide the interval \([0, 1]\) into subintervals and use the midpoint of each subinterval to evaluate the function. This method generally offers a better approximation than using left or right endpoints.
Let's break down the process of midpoint approximation:

• Divide the interval into \(n\) equal subintervals, each of width \(\Delta x\).
• The width \(\Delta x\) is given by \( \Delta x = \frac{b-a}{n} \).
• For each subinterval, calculate the midpoint \(m_i\) as \(m_i = \frac{x_i + x_{i-1}}{2}\).
• Evaluate the function \(f(x)\) at the midpoint \(m_i\).
• Compute the Riemann sum as the sum of the function values at the midpoints multiplied by \(\Delta x\): \( \sum_{i=1}^{n} f(m_i) \Delta x \).

The advantage of this approach is that as \(n\) increases, the approximation becomes more accurate, approaching the true value calculated by the definite integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of derivatives and integrals, showing that they are essentially inverse operations. This theorem is central to understanding why the process of integration works to calculate areas under curves. It consists of two parts:

• Part 1: It states that if a function \( f \) is continuous on the interval \([a, b]\) and \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
• This means we can find the exact value of a definite integral by simply evaluating the antiderivative at the upper and lower bounds and subtracting the results.
• Part 2: This part establishes that the derivative of an integral function \( G(x) = \int_{a}^{x} f(t) \, dt \) is \( G'(x) = f(x) \).

In the exercise, the use of this theorem allowed us to find the exact area under the curve \( f(x) = x^2 + 1 \) over the interval \([0, 1]\) by evaluating the antiderivative \( \frac{x^3}{3} + x \) at the limits. The Fundamental Theorem simplifies the process of solving many problems in calculus, providing a powerful tool for students.