91Ó°ÊÓ

When working with Riemann sums, the first step is to divide the total interval into smaller sections known as subintervals. In this exercise, we are given the interval \[0,1\] and need to divide it into 4 equal parts. The formula to calculate the width of each subinterval is \( \Delta x = \frac{b-a}{n} \), where \(a\) and \(b\) are the start and end of the interval, and \(n\) is the number of subintervals.
In this case:

• The total interval is \[0,1\]
• The number of subintervals, \(n\), is 4

So, the width of each subinterval is \( \Delta x = \frac{1-0}{4} = 0.25 \). This gives us the subintervals: \[0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1.0]\]. These subintervals form the base of the rectangles used in different Riemann sum methods.

The left-hand endpoint method is one way to approximate the area under a curve. With this approach, we use the left endpoint of each subinterval to determine the height of the rectangle. This means, for the subintervals \[0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1.0]\], we consider the points 0, 0.25, 0.5, and 0.75.
Here's a clearer look at what this means:

• \(f(0) = 0\)
• \(f(0.25) = -0.0625\)
• \(f(0.5) = -0.25\)
• \(f(0.75) = -0.5625\)

These values help construct the rectangles where each rectangle's height is the function value at the left endpoint of each subinterval. By multiplying these heights by \(\Delta x\), we approximate the integral as \(0 + (-0.0625) + (-0.25) + (-0.5625)) \times 0.25 = -0.21875\).
This method often underestimates the true area for functions that are decreasing across the interval.

The right-hand endpoint method uses the right endpoint of each subinterval to find the rectangle heights, offering another way to estimate the area under a curve. In our scenario for subintervals \[0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1.0]\], the focus shifts to 0.25, 0.5, 0.75, and 1.0.
Let's see how calculation changes:

• \(f(0.25) = -0.0625\)
• \(f(0.5) = -0.25\)
• \(f(0.75) = -0.5625\)
• \(f(1.0) = -1\)

These values set the height of each rectangle, aligning with the right endpoint of the subintervals. We then multiply the width by these values to compute the sum \( (-0.0625 + (-0.25) + (-0.5625) + (-1)) \times 0.25 = -0.46875 \).
The right-hand method can often overestimate the area for functions that are decreasing, providing a different perspective compared to the left-hand approach.

The midpoint method is slightly different because it uses the midpoint rather than the endpoints to determine the height of rectangles. This often provides a more accurate approximation compared to using just the left or right endpoints. For the subintervals \[0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1.0]\], the midpoints are 0.125, 0.375, 0.625, and 0.875.
Here are the calculations involved:

• \(f(0.125) = -0.015625\)
• \(f(0.375) = -0.140625\)
• \(f(0.625) = -0.390625\)
• \(f(0.875) = -0.765625\)

Each rectangle's height uses the value of the function at these midpoints, resulting in the sum \( (-0.015625 + (-0.140625) + (-0.390625) + (-0.765625)) \times 0.25 = -0.328125 \).
This method balances the estimates between under and over approximations, often resulting in a better estimate for the actual area under most function types.