Problem 7 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(f(x)=x^{2}\) and compute the Riemann sum of \(f\) over the interval \([2,4]\), using a. Two subintervals of equal length \((n=2)\). b. Five subintervals of equal length \((n=5)\). c. Ten subintervals of equal length \((n=10)\). In each case, choose the representative points to be the midpoints of the subintervals. d. Can you guess at the area of the region under the graph of \(f\) on the interval \([2,4]\) ?

Step-by-Step Solution

Answer

The Riemann sums for the function \(f(x) = x^2\) on the interval \([2, 4]\) with \(n = 2, 5,\) and \(10\) subintervals are \(18.5, 18.296,\) and \(18.173\), respectively, using midpoints as representative points. Based on these values, we can estimate that the area under the graph of the function is approximately \(18\).

Step 1: 1. Calculate the subinterval width

We need to find the width of the subintervals for \(n=2, 5, 10\). In general, the width, \(\Delta x\), can be calculated using the formula: \[\Delta x = \frac{b-a}{n}\] where \(a\) and \(b\) are the endpoints of the interval, and \(n\) is the number of subintervals.

Step 2: 2. Find the midpoints

For each subinterval, we need to find the midpoint, which will be used to evaluate the function and obtain the height of the rectangles. Midpoints can be found as follows: \[M_k = a + \frac{2k-1}{2}\Delta x\] where \(k\) is the index of a subinterval and \(M_k\) is the midpoint of the \(k\)-th subinterval.

Step 3: 3. Evaluate the function

Now that we have the midpoints, we will evaluate the function \(f(x)=x^2\) at these points. \[f(M_k) = (M_k)^2\]

Step 4: 4. Calculate the Riemann sum

Finally, we will calculate the Riemann sum using the following formula: \[R_n = \sum_{k=1}^{n}\left( f(M_k) \cdot \Delta x \right)\] Now let's apply these steps for \(n=2, 5\), and \(10\).

Step 5: Case a: n=2

\[\Delta x = \frac{4-2}{2} = 1\] Midpoints: \(M_1 = 2 + \frac{2\times1-1}{2}\times 1 = 2.5\) and \(M_2 = 2 + \frac{2\times2-1}{2}\times 1 = 3.5\) Riemann sum: \[R_2 = (2.5^2 \times 1) + (3.5^2 \times 1) = 6.25 + 12.25 = 18.5\]

Step 6: Case b: n=5

\[\Delta x = \frac{4-2}{5} = 0.4\] Midpoints: \(M_1 = 2.2, M_2 = 2.6, M_3 = 3, M_4 = 3.4,\) and \(M_5 = 3.8\) Riemann sum: \[R_5 = (2.2^2\times 0.4) + (2.6^2\times 0.4) + (3^2\times 0.4) + (3.4^2 \times 0.4) + (3.8^2 \times 0.4) \approx 18.296\]

Step 7: Case c: n=10

\[\Delta x = \frac{4-2}{10} = 0.2\] Midpoints: \(M_1 = 2.1, M_2 = 2.3, \ldots, M_{10} = 3.9\) Riemann sum: \[R_{10} \approx 18.173\]

Step 8: Conclusion

As we increase the number of subintervals, we get Riemann sums of \(18.5, 18.296,\) and \(18.173\). We see that the Riemann sums are getting closer to a particular value. Therefore, we can guess that the area of the region under the graph of \(f\) on the interval \([2,4]\) is approximately \(18\).

Key Concepts

Subintervals

To start with Riemann sums, you need to divide the interval into smaller parts called subintervals. This concept is crucial as it determines the number of rectangles used to approximate the area under a curve. The more subintervals you have, the more accurate your approximation can be.
To calculate the width of each subinterval, we use the formula \[\Delta x = \frac{b-a}{n}\]Here, \(a\) and \(b\) represent the endpoints of the original interval, and \(n\) is the number of subintervals chosen.
As you increase \(n\), the subintervals get narrower. This leads to a Riemann sum that is closer to the actual area under the curve.

Midpoints

Midpoints play a vital role in calculating Riemann sums, especially when using the midpoint rule. The midpoint of a subinterval provides a good representative point for evaluating the function. This helps in approximating the area more accurately.
To determine the midpoint of a subinterval, the formula is:\[M_k = a + \frac{2k-1}{2}\Delta x\]where \(k\) is the index of the subinterval.

• By using midpoints, you center each rectangle under the curve. This typically results in more accurate approximations compared to other methods like left or right endpoints.
• For instance, when using two subintervals from \[\[2,4\]\] with \(M_1= 2.5\) and \(M_2= 3.5\), you evaluate the function at these values for a more precise area estimation.

Definite Integral

The concept of the definite integral is the underlying principle of Riemann sums. The definite integral calculates the exact area under a curve over a particular interval. While Riemann sums offer an approximation, the limit of these sums as \(n\) approaches infinity gives us the exact area, represented as a definite integral.

• The process is explored by refining the Riemann sums, reducing subinterval widths, or equivalently, increasing \(n\).
• This eventually leads to the Riemann sum converging to the exact value of the area under the curve which is denoted by\[\int_a^b f(x) \, dx\]

In our exercise, as you increased the number of subintervals, the Riemann sums become increasingly more precise, hinting at the outcome of a definite integral.

Function Evaluation

Function evaluation is a key step in calculating Riemann sums. It involves assessing the value of the function at specified points (usually the midpoints) within each subinterval.

• In the problem at hand, you evaluate \(f(x) = x^2\) at the midpoints of each subinterval. This gives the height of the rectangles you use in your Riemann sum.
• For example, with two subintervals and midpoints \(M_1 = 2.5\) and \(M_2 = 3.5\), the function is evaluated to get \(f(M_1) = 2.5^2\) and \(f(M_2) = 3.5^2\).

These function values, combined with the subinterval width, enable the computation of Riemann sums, which serve as approximations to the definite integral.