Question

Sketch the graph of a continuous function on \((0,2)\) for which the right endpoint approximation with \(n = 2\) is more accurate than Simpson's Rule.

Short answer

The graph can be:

Step-by-step solution

Definition

Simpson's Rule\(\int_a^b f (x)dx \approx {S_n} = \frac{{\Delta x}}{3}\left( {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 4f\left( {{x_3}} \right) + L + 2f\left( {{x_{n - 2}}} \right) + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right)\)

Where\(x\)is even and\(\Delta x = \frac{{b - a}}{n}\).

Graph & its explanation

Simpson's rule is more accurate than the right endpoint approximation for most continuous functions (for all value of \(n\)), there are many different functions for those this will not be true.

Consider the graph:

This function \(f\) is a continuous function because there is no break or hole in it.

In this function, the right endpoint approximation with \(n = 2\) is more accurate than Simpson's Rule because the graph is very close to the two rectangles that the right endpoint approximation is based on & we know that the Simpson's rule is based on using parabolas to approximate the function, it will underestimate the area under the function.