Problem 3 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(f(x)=3 x\). a. Sketch the region \(R\) under the graph of \(f\) on the interval \([0,2]\) and find its exact area using geometry. b. Use a Riemann sum with four subintervals of equal length \((n=4)\) to approximate the area of \(R\). Choose the representative points to be the left endpoints of the subintervals. c. Repeat part (b) with eight subintervals of equal length \((n=8) .\) d. Compare the approximations obtained in parts (b) and (c) with the exact area found in part (a). Do the approximations improve with larger \(n\) ?

Step-by-Step Solution

Answer

In summary, the exact area under the graph of \(f(x)=3x\) on the interval \([0,2]\) is 6 square units. Approximating the area using Riemann sums with \(n=4\) subintervals gives a result of 4.5 square units, while using \(n=8\) subintervals yields an approximation of 5.25 square units. The approximations improve with a larger number of subintervals, as they better capture the graph's behavior.

Step 1: a. Sketch the region R and find its exact area using geometry.

We're looking for the area under the graph of \(f(x) = 3x\) on the interval \([0, 2]\). First, we need to find the points where the function intersects the interval endpoints. At \(x=0\), we have \(f(0) = 3(0) = 0\), and at \(x=2\), we have \(f(2) = 3(2) = 6\). This means, the region R starts from the point \((0, 0)\) and ends at the point \((2, 6)\), and makes a right triangle with the x-axis and y-axis. To find the exact area of this triangle, use the formula for the area of a triangle: \(\frac{1}{2} \cdot \text{base} \cdot \text{height}\), which in this case is \(\frac{1}{2}(2)(6) = 6\).

Step 2: b. Use a Riemann sum with n=4 subintervals to approximate the area of R.

To use a Riemann sum with four subintervals of equal length, we first need to divide the interval \([0, 2]\) into four equal parts: \([0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]\). Now, choosing the left endpoints of the subintervals as representative points, we have \(x_1 = 0, x_2 = 0.5, x_3 = 1\), and \(x_4 = 1.5\). Then, the Riemann sum with four subintervals (of equal length \(\Delta x = 0.5\)) is given by: \[R_4 = \sum_{i=1}^{4} f(x_i) \Delta x = \sum_{i=1}^{4} 3x_i \Delta x = \sum_{i=1}^{4} 3x_i(0.5)\] So, calculate the Riemann sum: \[R_4 = 3(0)(0.5) + 3(0.5)(0.5) + 3(1)(0.5) + 3(1.5)(0.5) = 0 + 0.75 + 1.5 + 2.25 = 4.5\]

Step 3: c. Repeat part (b) with n=8 subintervals.

With eight subintervals of equal length, we divide the interval \([0, 2]\) into eight equal parts: \([0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1], [1, 1.25], [1.25, 1.5], [1.5, 1.75], [1.75, 2]\). Choosing the left endpoints of the subintervals, we have \(x_1 = 0, x_2 = 0.25, x_3 = 0.5, x_4 = 0.75, x_5 = 1, x_6 = 1.25, x_7 = 1.5\), and \(x_8 = 1.75\). Then, the Riemann sum with eight subintervals (of equal length \(\Delta x = 0.25\)) is given by: \[R_8 = \sum_{i=1}^{8} f(x_i) \Delta x = \sum_{i=1}^{8} 3x_i \Delta x = \sum_{i=1}^{8} 3x_i(0.25)\] So, calculate the Riemann sum: \[R_8 = 3(0)(0.25) + 3(0.25)(0.25) + 3(0.5)(0.25) + 3(0.75)(0.25) + 3(1)(0.25) + 3(1.25)(0.25) + 3(1.5)(0.25) + 3(1.75)(0.25)\] \[R_8 = 0 + 0.1875 + 0.375 + 0.5625 + 0.75 + 0.9375 + 1.125 + 1.3125 = 5.25\]

Step 4: d. Compare the approximations obtained in parts (b) and (c) with the exact area found in part (a). Do the approximations improve with larger n?

In part (a), we found the exact area under the graph of \(f(x) = 3x\) on the interval \([0, 2]\) to be 6 square units. Using Riemann sums with \(n=4\) subintervals, we calculated an approximate area of 4.5 square units, and with \(n=8\) subintervals, we calculated an approximate area of 5.25 square units. Both approximations are smaller than the exact area, but the approximation with more subintervals (n=8) is closer to the exact area. This demonstrates that the approximation does improve with larger n, as the subintervals get smaller and more representative of the graph's behavior.

Key Concepts

Definite Integrals

Definite integrals offer a way to calculate the exact area under a curve from one point to another on a graph. When we speak about a definite integral \( \int_{a}^{b} f(x) \, dx \), we're describing the accumulated area between the curve and the horizontal axis from \( x = a \) to \( x = b \). Understanding definite integrals revolves around establishing the relationship between the integral and the geometric concept of area.

• Place the function on a coordinate plane.
• Identify the interval over which you solve, such as \([0, 2]\) in our example.
• Use the function's definition within that range to compute the \( y \)-values, bounding the area.

This method allows us to precisely find areas that might otherwise be complex or irregular shapes.
It applies especially well to problems like determining the exact area under the function \( f(x) = 3x \) from \( 0 \) to \( 2 \) using geometric shapes such as triangles.

Area under a Curve

Finding the area under a curve involves calculating the space the curve occupies when plotted against the horizontal axis. The challenge lies in accurately representing this area regardless of the curve's shape.The curve of a function like \( y = 3x \) over an interval such as \( [0, 2] \) contributes directly to understanding the region it forms with the \( x \)-axis and vertical lines at the chosen interval endpoints.

• The area significance relates to real-world applications such as finding distances or quantities.
• A geometric approach, like using a triangle in this instance, is a simple form to visualize and solve for area.
• The formula for the area of a triangle \( \left( \frac{1}{2} \text{base} \times \text{height} \right) \) applies effectively to linear functions.

By recognizing the importance of these calculations, you can gain deeper insights into functional behaviors in physics, economics, and more.

Approximation Methods

Approximation methods such as Riemann sums provide ways to estimate the areas under curves when exact calculations prove challenging.In our example, the Riemann sum is done by:\

• Dividing the total interval into smaller equal sections called subintervals.
• Calculating the function's value at chosen points within each subinterval (in this case, the left endpoints).
• Summing the products of these values and the subinterval width.

For example, splitting \( [0, 2] \) into four parts gives a rougher approximation, while eight parts provide a more refined estimate.
Refinement comes from how these smaller interval sections mirror more closely the curve's subtle changes, increasing approximation accuracy.

Subintervals in Integration

Breaking down the interval over which you're integrating into subintervals forms a core part of many approximation techniques like Riemann sums. Each subinterval provides a slice of the whole, and many such slices combine to approach the complete area.

• Subintervals can vary in number and length based on the required approximation precision.
• The width of each subinterval \( \Delta x \) directly impacts the closeness of the approximation.
• More subintervals with smaller widths typically lead to higher accuracy.

For instance, our exercise used both four and eight subintervals: the latter offered a finer approximation.
This part of definite integral approximation helps students understand the benefits of different partition strategies, showing how calculus can describe the real world more and more accurately as these divisions become finer.