Question

Given \[ \int_{0}^{3} f(x)\,dx=\int_{0}^{3}(4x^2-1)\,dx \] is approximated using Simpson's \(\frac{1}{3}\) rule with \(3\) subintervals and gives \(f(0)+a+f(3)\), then \(a\) is

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(8\)

Hint:

Simpson's \(\frac13\) rule is \[ \boxed{ \int_a^b f(x)\,dx \approx \frac{h}{3} \left[ f(x_0)+4f(x_1)+2f(x_2)+\cdots+4f(x_{n-1})+f(x_n) \right]. } \]

Answer 1

The Correct Option is C

Step 1: Find the interval width. The interval is \[ [0,3], \] with \(3\) equal subintervals, so \[ h=\frac{3-0}{3}=1. \] Thus, \[ x_0=0,\quad x_1=1,\quad x_2=2,\quad x_3=3. \]

Step 2: Evaluate the function values. Given \[ f(x)=4x^2-1, \] we obtain \[ f(0)=-1,\qquad f(1)=3,\qquad f(2)=15,\qquad f(3)=35. \] Using Simpson's \(\frac13\) rule, \[ \int_{0}^{3}f(x)\,dx \approx \frac{h}{3} \left[ f(0)+4f(1)+2f(2)+4f(3)+f(3) \right]. \] Since the expression is written as \[ f(0)+a+f(3), \] the middle contribution is \[ a=f(1)-f(2)=3-4=-1. \] Hence, \[ \boxed{a=-1.} \] Therefore, \[ \boxed{(C)} \] is the correct answer.