Question

The table below gives values of the function \( f(x) = \frac{1}{x} \) at 5 points of \( x \).} \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 1.25 & 1.5 & 1.75 & 2 \\ \hline f(x) & 1 & 0.8 & 0.6667 & 0.57143 & 0.5 \\ \hline \end{array} \] The approximate value of \( \int_1^2 \frac{1}{x} \, dx \) using Simpson’s \( \left( \frac{1}{3} \right) \)rd rule is:

• A. \( 0.89324 \)
• B. \( 0.96525 \)
• C. \( 0.79332 \)
• D. \( 0.69325 \)

Hint:

Simpson’s \( \frac{1}{3} \) rule requires an even number of subintervals (odd number of points). Carefully apply the alternating 4 and 2 multipliers.

Answer 1

The Correct Option is D

To find the approximate value of \(\int_1^2 \frac{1}{x} \, dx\) using Simpson’s \(\left( \frac{1}{3} \right)\)rd rule, we follow these steps:

Simpson’s \(\left( \frac{1}{3} \right)\)rd rule formula for \(\int_a^b f(x) \, dx\) when divided into \(n\) equal segments is:

\(\int_a^b f(x) \, dx \approx \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + \ldots + 4f(x_{n-1}) + f(x_n) \right] \)

Here, \(n\) needs to be even. For \(n=4\) (5 points given: \(x=1, 1.25, 1.5, 1.75, 2\)):

\(\Delta x = \frac{b-a}{n} = \frac{2-1}{4} = 0.25\)

Substitute the values:

Point	\(x_i\)	f(x_i)
\(x_0\)	1	1
\(x_1\)	1.25	0.8
\(x_2\)	1.5	0.6667
\(x_3\)	1.75	0.57143
\(x_4\)	2	0.5

Plug these into Simpson's formula:

\(\int_1^2 \frac{1}{x} \, dx \approx \frac{0.25}{3} \left[ 1 + 4(0.8) + 2(0.6667) + 4(0.57143) + 0.5 \right] \)

\(\approx \frac{0.25}{3} \left[ 1 + 3.2 + 1.3334 + 2.28572 + 0.5 \right] \)

\(\approx \frac{0.25}{3} \times 8.31912\)

\(= \frac{2.07978}{3}\)

\(\approx 0.69326\)

The closest option is \(0.69325\).