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 use Simpson’s \( \frac{1}{3} \) rule, we apply: \[ \int_a^b f(x)\,dx \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \dots + f(x_n) \right] \] Here, 5 points mean \( n = 4 \) (even), so \( h = \frac{2 - 1}{4} = 0.25 \) \[ \begin{aligned} \int_1^2 \frac{1}{x} dx &\approx \frac{0.25}{3} [1 + 4(0.8) + 2(0.6667) + 4(0.57143) + 0.5] \\ &= \frac{0.25}{3} [1 + 3.2 + 1.3334 + 2.28572 + 0.5] \\ &= \frac{0.25}{3} \cdot 8.31912 \\ &\approx 0.69325 \end{aligned} \]