Problem 50 - Calculus Volume 1

Question

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums \(R_{N}\) and \(L_{N}\) for \(N=1,10,100\) . How do these estimates compare with the exact answers, which you can find via geometry? \([T] \mathrm{y}=3 x+2\) on the interval \([3,5]\)

Step-by-Step Solution

Answer

Endpoint sums approximate the exact area of 28, with accuracy increasing for larger N.

Step 1: Understand the Problem

We need to find the right endpoint sum (\(R_N\)) and left endpoint sum (\(L_N\)) for different values of \(N\) (1, 10, and 100) on the interval [3,5] for the function \(y=3x+2\). We'll compare these sums with the exact area found using geometric methods.

Step 2: Calculate the Right Endpoint Sums (R_N)

For each \(N\), calculate the width \(\Delta x = \frac{5-3}{N}\). The right endpoints for each sub-interval will be \(x_i = 3 + i\Delta x\), where \(i = 1, 2, ..., N\). Sum the heights of the rectangles \(R_N=\sum_{i=1}^{N} [3(3+i\Delta x)+2] \Delta x\).

Step 3: Substep: Example Calculation of \(R_1\)

With \(N=1\), the width is \(\Delta x=2\). The right endpoint is at \(x=5\). The height at this point is \(f(5)=3(5) + 2 = 17\). Thus, \(R_1 = 17(2) = 34\).

Step 4: Substep: Example Calculation of \(R_{10}\)

With \(N=10\), \(\Delta x=0.2\). Calculate heights at each endpoint: from \(3.2\) to \(5\) in steps of \(0.2\). Sum these heights times \(0.2\) to find \(R_{10}\).

Step 5: Substep: Example Calculation of \(R_{100}\)

With \(N=100\), \(\Delta x=0.02\). Calculate heights at each endpoint: from \(3.02\) to \(5\) in steps of \(0.02\). Sum these heights times \(0.02\) to find \(R_{100}\).

Step 3: Calculate the Left Endpoint Sums (L_N)

For each \(N\), use similar calculations as \(R_N\), except starting at the left of each sub-interval \(x_i = 3 + (i-1)\Delta x\). Sum the heights of rectangles \(L_N=\sum_{i=0}^{N-1} [3(3+i\Delta x)+2] \Delta x\).

Step 7: Substep: Example Calculation of \(L_1\)

With \(N=1\), the width is \(\Delta x=2\). The left endpoint is at \(x=3\). The height at this point is \(f(3)=3(3) + 2 = 11\). Thus, \(L_1 = 11(2) = 22\).

Step 8: Substep: Example Calculation of \(L_{10}\)

With \(N=10\), \(\Delta x=0.2\). Calculate heights at each endpoint: from \(3\) to \(4.8\) in steps of \(0.2\). Sum these heights times \(0.2\) to find \(L_{10}\).

Step 9: Substep: Example Calculation of \(L_{100}\)

With \(N=100\), \(\Delta x=0.02\). Calculate heights at each endpoint: from \(3\) to \(4.98\) in steps of \(0.02\). Sum these heights times \(0.02\) to find \(L_{100}\).

Step 4: Compare with the Exact Area

The area under the line \(y=3x+2\) from \(x=3\) to \(x=5\) can also be found using the formula for areas under straight line segments. This region is a trapezoid with bases \(y(3)=11\) and \(y(5)=17\), and height \(2\), giving an area of \((11+17)/2 \times 2 = 28\).

Step 11: Conclusion: Comparison and Understanding

As \(N\) increases, both \(R_N\) and \(L_N\) approach the exact area of 28. Initially, \(L_N\) underestimates and \(R_N\) overestimates the area, which is typical for lower subdivision numbers.

Key Concepts

Right Endpoint Approximation

When approximating the area under a curve using right endpoint approximation, we imagine dividing the interval into smaller widths known as sub-intervals. For each sub-interval, we take the value of the function at the right endpoint and use it as the height of our rectangle. The base of the rectangle is the width of each sub-interval, denoted as \( \Delta x \).
The right endpoint approximation is represented mathematically as follows:

• Calculate the width of each sub-interval: \( \Delta x = \frac{b-a}{N} \)
• Determine the right endpoint for each rectangle: \( x_i = a + i \Delta x \) for \( i=1,2,...,N \)
• Calculate the height of each rectangle using the function at those points: \( f(x_i) \)
• Formulate the right endpoint sum: \( R_N = \sum_{i=1}^{N} f(x_i) \Delta x \)

This method often leads to an overestimation of the area under the curve if the function is increasing, as it was with the function \( y=3x+2 \) on the interval [3,5]. This is evident with \( R_1 = 34 \), which is an overestimate compared to the geometric solution of 28.

Left Endpoint Approximation

In left endpoint approximation, we use the value of the function at the left endpoint of each sub-interval to estimate the area under a curve. This involves creating rectangles where the height is determined by the leftmost point of the sub-interval, and again, each rectangle's base is \( \Delta x \).
The process is mathematically described as follows:

• Determine the width of each sub-interval, \( \Delta x = \frac{b-a}{N} \)
• Identify the left endpoint for each rectangle: \( x_i = a + (i-1) \Delta x \) for \( i=0,1,...,N-1 \)
• Use these points to find the height of each rectangle: \( f(x_i) \)
• Compose the left endpoint sum as: \( L_N = \sum_{i=0}^{N-1} f(x_i) \Delta x \)

This approximation tends to underestimate the area for increasing functions. With our example, we see \( L_1 = 22 \), falling short of the exact geometric area of 28, demonstrating an initial underestimate.

Geometric Area Calculation

Calculating the area under a line geometrically can sometimes give the exact value, avoiding the approximate nature of endpoint sums. For a linear function, this can be as simple as finding the area of basic geometric shapes such as rectangles, triangles, or trapezoids. In the example function \( y = 3x + 2 \), the interval from [3,5] is a straight line, forming a trapezoid.
To find the area of a trapezoid:

• Determine the lengths of the two parallel sides (the bases) by evaluating the function at the endpoints: \( y(3) = 11 \), \( y(5) = 17 \)
• The height of the trapezoid is simply the difference in the x-values: \( 5-3 = 2 \)
• Use the formula for the area of a trapezoid: \( \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \)
• Substitute the values: \( \frac{1}{2} \times (11 + 17) \times 2 = 28 \)

Hence, this geometric method provides an exact area of 28, illustrating that as the number of sub-intervals in endpoint sums increases, their estimates approximate the exact geometric area more closely.