Question

The values of \(x\) and \(f(x)\) of a curve are as follows: \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 2 & 2.5 & 3 & 3.5 & 4 \hline f(x) & 3 & 6 & 8 & 10 & 12 \hline \end{array} \] By Simpson's one-third rule, the area under the curve \(f(x)\) is ____.

• A. \(26.3\)
• B. \(31.7\)
• C. \(13.2\)
• D. \(15.8\)

Hint:

Simpson's one-third rule: \[ \boxed{ \int_a^b f(x)\,dx = \frac{h}{3} \left[ y_0+y_n +4(\text{odd terms}) +2(\text{even terms}) \right] } \] It requires an even number of subintervals.

Answer 1

The Correct Option is D

Concept: Simpson's one-third rule is \[ \boxed{ \int_a^b f(x)\,dx = \frac{h}{3} \left[ y_0+y_n +4(y_1+y_3+\cdots) +2(y_2+y_4+\cdots) \right] } \] where \(h\) is the equal interval.

Step 1: Identify the data. \[ h=2.5-2=0.5 \] The function values are \[ y_0=3,\; y_1=6,\; y_2=8,\; y_3=10,\; y_4=12. \]

Step 2: Apply Simpson's one-third rule. \[ \int_2^4 f(x)\,dx = \frac{0.5}{3} \left[ 3+12+4(6+10)+2(8) \right] \] \[ = \frac{0.5}{3} \left[ 15+64+16 \right] \] \[ = \frac{0.5}{3}\times95 = 15.83. \] Hence, \[ \boxed{\int_2^4 f(x)\,dx\approx15.8.} \] Therefore, the correct option is \[ \boxed{(D)\;15.8.} \]