Problem 5 - Calculus: Early Transcendentals

Question

(a) Estimate the area under the graph of \( f(x) = 1 + x^2 \) from \( x = -1 \) to \( x = 2 \) using three rectangles and right endpoints. Then improve your estimate by using six rectangles. Sketch the curve and the approximating rectangles. (b) Repeat part (a) using left endpoints. (c) Repeat part (a) using midpoints. (d) From your sketches in parts (a)-(c), which appears to be the best estimate?

Step-by-Step Solution

Answer

Using midpoints with six rectangles (6.4375) gives the best estimate.

Step 1: Estimate using three rectangles with right endpoints

We divide the interval from \(x = -1\) to \(x = 2\) into three subintervals: \([-1, 0]\), \([0, 1]\), and \([1, 2]\). The width of each rectangle is \(\Delta x = 1\).The right endpoints for these intervals are \(x = 0\), \(x = 1\), and \(x = 2\).For each rectangle:\[ f(0) = 1 + 0^2 = 1 \]\[ f(1) = 1 + 1^2 = 2 \]\[ f(2) = 1 + 2^2 = 5 \]Thus, the area estimate is:\[ \text{Area} = \Delta x \times (f(0) + f(1) + f(2)) = 1 \times (1 + 2 + 5) = 8 \]

Step 2: Estimate using six rectangles with right endpoints

Divide the interval \([-1, 2] \) into six equal subintervals: \([-1, -0.5]\), \([-0.5, 0]\), \([0, 0.5]\), \([0.5, 1]\), \([1, 1.5]\), \([1.5, 2]\). The width of each rectangle is \(\Delta x = 0.5\).The right endpoints of these subintervals are: \(-0.5, 0, 0.5, 1, 1.5, 2 \).For each rectangle:\[ f(-0.5) = 1 + (-0.5)^2 = 1.25 \]\[ f(0) = 1 \]\[ f(0.5) = 1 + 0.5^2 = 1.25 \]\[ f(1) = 2 \]\[ f(1.5) = 1 + 1.5^2 = 3.25 \]\[ f(2) = 5 \]The area estimate is:\[ \text{Area} = \Delta x \times (f(-0.5) + f(0) + f(0.5) + f(1) + f(1.5) + f(2)) = 0.5 \times (1.25 + 1 + 1.25 + 2 + 3.25 + 5) = 6.875 \]

Step 3: Estimate using three rectangles with left endpoints

Divide the interval from \(x = -1\) to \(x = 2\) into three subintervals: \([-1, 0]\), \([0, 1]\), and \([1, 2]\), with \(\Delta x = 1\).Using the left endpoints, \(-1, 0, 1\):\[ f(-1) = 1 + (-1)^2 = 2 \]\[ f(0) = 1 \]\[ f(1) = 2 \]Area:\[ \text{Area} = \Delta x \times (f(-1) + f(0) + f(1)) = 1 \times (2 + 1 + 2) = 5 \]

Step 4: Estimate using six rectangles with left endpoints

Divide the interval \([-1, 2] \) into six subintervals: \([-1, -0.5]\), \([-0.5, 0]\), \([0, 0.5]\), \([0.5, 1]\), \([1, 1.5]\), \([1.5, 2]\), each of width \(\Delta x = 0.5\).Left endpoints are: \(-1, -0.5, 0, 0.5, 1, 1.5 \).For each rectangle:\[ f(-1) = 2 \]\[ f(-0.5) = 1.25 \]\[ f(0) = 1 \]\[ f(0.5) = 1.25 \]\[ f(1) = 2 \]\[ f(1.5) = 3.25 \]The area estimate:\[ \text{Area} = \Delta x \times (f(-1) + f(-0.5) + f(0) + f(0.5) + f(1) + f(1.5)) = 0.5 \times (2 + 1.25 + 1 + 1.25 + 2 + 3.25) = 5.875 \]

Step 5: Estimate using three rectangles with midpoints

Divide the interval into three subintervals: \([-1, 0]\), \([0, 1]\), and \([1, 2]\), \(\Delta x=1\).Use midpoints: \(-0.5, 0.5, 1.5\) \[ f(-0.5) = 1.25 \]\[ f(0.5) = 1.25 \]\[ f(1.5) = 3.25 \]Area:\[ \text{Area} = \Delta x \times (f(-0.5) + f(0.5) + f(1.5)) = 1 \times (1.25 + 1.25 + 3.25) = 5.75 \]

Step 6: Estimate using six rectangles with midpoints

Divide the interval \([-1, 2] \) into six subintervals: \([-1, -0.5]\), \([-0.5, 0]\), \([0, 0.5]\), \([0.5, 1]\), \([1, 1.5]\), \([1.5, 2]\), each with \(\Delta x = 0.5\).Midpoints are: \(-0.75, -0.25, 0.25, 0.75, 1.25, 1.75\).For each rectangle:\[ f(-0.75) = 1 + (-0.75)^2 = 1.5625 \]\[ f(-0.25) = 1.0625 \]\[ f(0.25) = 1.0625 \]\[ f(0.75) = 1.5625 \]\[ f(1.25) = 2.5625 \]\[ f(1.75) = 4.0625 \]The area estimate:\[ \text{Area} = \Delta x \times (f(-0.75) + f(-0.25) + f(0.25) + f(0.75) + f(1.25) + f(1.75)) = 0.5 \times (1.5625 + 1.0625 + 1.0625 + 1.5625 + 2.5625 + 4.0625) = 6.4375 \]

Step 7: Determine the best estimate

Comparing the estimates: - Right endpoint (3 rectangles): Area = 8 - Right endpoint (6 rectangles): Area = 6.875 - Left endpoint (3 rectangles): Area = 5 - Left endpoint (6 rectangles): Area = 5.875 - Midpoint (3 rectangles): Area = 5.75 - Midpoint (6 rectangles): Area = 6.4375 The estimate using midpoints with six rectangles (6.4375) seems to be the best estimate due to midpoint rule typically being more accurate than left or right endpoint methods.

Key Concepts

Rectangular Approximation

Numerical integration is a crucial concept in mathematics when dealing with the estimation of areas under curves, especially for functions that are hard or impossible to integrate analytically. One simple yet significant technique is a rectangular approximation.

This method involves subdividing the area under the curve into rectangles. The total area of these rectangles gives us an approximation of the area under the curve. Depending on how we define the height of each rectangle, there are different ways to implement this method. Through these approximations, we achieve a better grasp of how integral calculus applies practically. The focus on rectangles simplifies the integral's complexity, making it a favored introductory method for students learning numerical integration.

Importantly, the accuracy of this method greatly depends on the number and width of rectangles used. More rectangles typically lead to a more precise approximation because they can better conform to the curves. However, it can become computationally demanding as the number of rectangles increases.

Right Endpoint Approximation

Right endpoint approximation is one specific approach to rectangular approximation.

With this method, each rectangle's height is determined by the value of the function at the right endpoint of the subinterval. This means the function's value at the end of the subinterval dictates how tall the rectangle is. Thus, if you're considering a subinterval from Point A to Point B, the height is calculated at Point B.

• If the function is increasing on the interval, this method might overestimate the area because the rectangle is taller than required.
• Conversely, if the function is decreasing, it might underestimate the area because the rectangle will be shorter than the average height of the curve over that interval.

While simple, it's essential to recognize the limitations of right endpoint approximations when estimating areas. It's most effective when fine-tuning involves increasing the count of the subintervals, which diminishes the error due to both over and underestimation.

Left Endpoint Approximation

Left endpoint approximation is quite similar to its right endpoint counterpart but employs the value of the function at the left endpoint of each subinterval to define a rectangle's height.

For a subinterval from Point A to Point B, the height is set at Point A. This has distinct implications:

• For increasing functions, this method might underestimate the area, as the rectangles do not reach the full height of the curve in many parts.
• Conversely, for decreasing functions, it can overestimate because the start of the interval gives a taller rectangle than needed.

Just like other endpoint methods, its precision increases with more rectangles as it allows for better coverage and alignment with the curve. This method gives students another simple but powerful tool for understanding numerical integration, illustrating how different endpoint selections influence area estimates.

Midpoint Rule

The midpoint rule stands out because of its often improved approximation over left and right endpoint methods. Here, the rectangle's height is determined by the value of the function at the midpoint of the subinterval.

Suppose you have a subinterval running from Point A to Point B. In that case, the height of the rectangle for this interval will be calculated at the midpoint, \(C = \frac{A + B}{2}\). This strategy usually offers a more balanced approximation since:

• It accounts for the average height of the function over the subinterval rather than just one end, mitigating the errors associated with the other two endpoint methods.
• The midpoint rule can handle curves with more fluctuations more effectively, as it naturally accommodates changes within the subintervals.

Though this method can require additional calculation for the midpoints, it's fundamentally an advantageous approach when higher accuracy is desired from fewer subdivisions. Many students find this method appealing as it frequently yields a more accurate area approximation with less effort when the same number of rectangles is used.