Question

Use the function values in the following table and the Trapezoidal Rule with \(n=6\) to approximate \(\int_{0}^{6} f(x) d x\) $$\begin{array}{c|cccccc}{x} & {0} & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {12} & {10} & {9} & {11} & {13} & {16} & {18}\end{array}$$

Short answer

The approximation of \(\int_{0}^{6} f(x) d x\) using the Trapezoidal Rule with \(n=6\) is 133.

Step-by-step solution

Calculate h

The first step is to calculate the value of \(h\), which is the width of the intervals. From the exercise, \(a = 0\), \(b = 6\), and \(n = 6\). So, the width of the intervals \(h = \frac{b-a}{n} = \frac{6-0}{6} = 1\)

Identify the function values

The table provides the function values \(f(x)\) at the points \(x = 0, 1, 2, 3, 4, 5, 6\). These are \(f(x) = 12, 10, 9, 11, 13, 16, 18\) respectively.

Implement the Trapezoidal Rule

Using the formula for the Trapezoidal Rule, the approximated integral is computed as follows: \(\int_{0}^{6} f(x) d x \approx h \[\frac{f(a) + f(b)}{2} + 2 * \sum_{i=1}^{n-1} f(a + i * h)\] = 1 * \[\frac{f(0) + f(6)}{2} + 2 * \sum_{i=1}^{5} f(i)\] = \frac{12 + 18}{2} + 2 * (10 + 9 + 11 + 13 + 16) = 15 + 118 = 133\)