Question

Sketch the graph of a continuous function on\((0,2)\) for which the Trapezoidal Rule with\(n = 2\)is more accurate than the Midpoint Rule.

Short answer

The curve is\(f(x) = \sin x\).

Step-by-step solution

Definition

Trapezoid Rule

If\((a,b)\)is divided into\(n\)subintervals, each of length\(\Delta x\), with endpoints at\(\left\{ {{x_0},{x_1},{x_2}, \ldots ,{x_n}} \right\}\), then\({T_n} = \frac{{\Delta x}}{2}\left( {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + \cdots + 2f\left( {{x_{n - 2}}} \right) + 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right){\rm{ }}\)

Midpoint Rule

If\((a,b)\)is divided into\(n\)subintervals, each of length\(\Delta x\)and\({m_i}\)is the midpoint of the\(i\)th subinterval, then\({M_n} = \sum\limits_{i = 1}^n f \left( {{m_i}} \right)\Delta x\).

For Trapezoidal rule

Consider the continuous function\(f(x) = {x^2}\)on the interval\((0,2)\)

Area under the curve \(f(x) = \sin x\) using Trapezoidal rule is given by the curve below:

For Midpoint rule

Area under the curve\(f(x) = \sin x\) using Midpoint rule is given below:

From the above diagrams, loss of area by Midpoint Rule is more than that of Trapezoidal rule. That is area by Trapezoidal is accurate than that of Midpoint rule as trapezoidal rule works on trapezoids.