91Ó°ÊÓ

The left endpoint approximation is a simple yet effective technique used to estimate the area under a curve. This method involves evaluating the function at the beginning, or left end, of each sub-interval. Since we are working over the interval \([0,5]\) with 5 sub-intervals, each sub-interval here has a width of \( \Delta x = 1 \).

To perform the left endpoint approximation, you calculate the value of the function at each left endpoint within the range. For the function \( y = x^2 + 1 \), this means evaluating at \( x = 0, 1, 2, 3, \) and \( 4 \).

Here's a breakdown:

• \( f(0) = 1 \)
• \( f(1) = 2 \)
• \( f(2) = 5 \)
• \( f(3) = 10 \)
• \( f(4) = 17 \)

You then multiply these function values by the width of the sub-interval \( \Delta x = 1 \) and sum them up. This gives us:\[ A_L = 1 \times (1 + 2 + 5 + 10 + 17) = 35 \]This value represents the estimated area using left endpoint approximation.

The right endpoint approximation works similarly to the left endpoint, but instead evaluates the function at the end of each sub-interval. In this method, for each interval, we consider the right-most x-value when calculating the area.

Just like the left endpoint, the sub-interval width is \( \Delta x = 1 \). This time, with the same function \( y = x^2 + 1 \), we check the values at points \( x = 1, 2, 3, 4, \) and \( 5 \) to include each endpoint. This spans from the first point after 0 until the boundary at 5.

Function evaluations are:

• \( f(1) = 2 \)
• \( f(2) = 5 \)
• \( f(3) = 10 \)
• \( f(4) = 17 \)
• \( f(5) = 26 \)

These are then multiplied by the width of each sub-interval and added up, calculating the approximation:\[ A_R = 1 \times (2 + 5 + 10 + 17 + 26) = 60 \]The result, 60, serves as the right endpoint approximation of the area under the curve.

The midpoint approximation provides a compelling alternative by averaging values over each sub-interval for a potentially more accurate area estimation under the curve. Here, the function is evaluated at the midpoints of each sub-interval.

Given the same interval \([0, 5]\) and function \( y = x^2 + 1 \), we use the midpoints \( x = 0.5, 1.5, 2.5, 3.5, \) and \( 4.5 \). Since each midpoint is central to a sub-interval of width \( \Delta x = 1 \), only these need to be evaluated to find our approximation.

By computing these midpoint values, we have:

• \( f(0.5) = 1.25 \)
• \( f(1.5) = 3.25 \)
• \( f(2.5) = 7.25 \)
• \( f(3.5) = 13.25 \)
• \( f(4.5) = 21.25 \)

Then multiplying by the interval width and summing the values provides the estimated area:\[ A_M = 1 \times (1.25 + 3.25 + 7.25 + 13.25 + 21.25) = 46.25 \]This midpoint approximation can often be closer to the true area than simple endpoint methods.