Question

If $f(x)$ is defined on $[a,b]$ and $[a,b]$ is divided into $2n$ equal parts by the points $x_0=a<x_1<x_2<...<x_{2n}=b$. Then by Simpson's $\frac{1}{3}$ rule $\int_a^b f(x)dx =$

• A. $\frac{b-a}{3n} [(f(x_0)+f(x_{2n}))+4(f(x_1)+f(x_3)+\dots+f(x_{2n-1}))+2(f(x_2)+f(x_4)+\dots+f(x_{2n-2}))]$
• B. $\frac{b-a}{6n} [(f(x_0)+f(x_{2n}))+4(f(x_1)+f(x_3)+\dots+f(x_{2n-1}))+2(f(x_2)+f(x_4)+\dots+f(x_{2n-2}))]$
• C. $\frac{b-a}{6n} [(f(x_0)+f(x_{2n}))+4(f(x_1)+f(x_3)+\dots+f(x_{2n-1}))+2(f(x_2)+f(x_4)+\dots+f(x_{2n-2}))]$
• D. $\frac{b-a}{3n} [(f(x_0)+f(x_{2n}))+4(f(x_1)+f(x_3)+\dots+f(x_{2n-2}))+2(f(x_2)+f(x_4)+\dots+f(x_{2n-1}))]$

Hint:

• Simpson's $\frac{1}{3}$ rule requires an even number of subintervals ($2n$).
• Width of each subinterval $h = (b-a)/(2n)$.
• Formula: $\int_a^b f(x)dx \approx \frac{h}{3} [y_0 + y_{2n} + 4(y_1+y_3+\dots+y_{2n-1}) + 2(y_2+y_4+\dots+y_{2n-2})]$.
• Substitute $h = (b-a)/(2n)$ to get the coefficient $\frac{b-a}{6n}$.
• Pattern of coefficients for ordinates: 1, 4, 2, 4, 2, ..., 4, 2, 4, 1.

Answer 1

The Correct Option is C

We are given a function \( f(x) \) defined on the interval \([a,b]\). The interval \([a,b]\) is divided into \(2n\) equal subintervals by the points:

\[ x_0 = a<x_1<x_2<\dots<x_{2n} = b \]

We want to approximate the definite integral:

\[ \int_a^b f(x) \, dx \]

Step 1: Define the width of each subinterval

Since the interval \([a,b]\) is divided into \(2n\) equal parts, the width of each small subinterval \(\Delta x\) is:

\[ \Delta x = \frac{b - a}{2n} \]

Step 2: Simpson's \(\frac{1}{3}\) Rule for \(2n\) subintervals

Simpson's rule approximates the integral by fitting parabolas through pairs of subintervals. For \(2n\) subintervals, the formula is:

\[ \int_a^b f(x) \, dx \approx \frac{\Delta x}{3} \left[ f(x_0) + 4 f(x_1) + 2 f(x_2) + 4 f(x_3) + 2 f(x_4) + \cdots + 4 f(x_{2n-1}) + f(x_{2n}) \right] \]

Step 3: Group terms according to their coefficients

• First and last terms: \( f(x_0) \) and \( f(x_{2n}) \), each counted once.
• Terms with odd indices (i.e., \(x_1, x_3, \ldots, x_{2n-1}\)) are multiplied by 4.
• Terms with even indices except the first and last (i.e., \(x_2, x_4, \ldots, x_{2n-2}\)) are multiplied by 2.

Step 4: Substitute \(\Delta x = \frac{b - a}{2n}\) in the formula

\[ \int_a^b f(x) \, dx \approx \frac{b - a}{6n} \left[ f(x_0) + 4 \sum_{i=1, \text{odd}}^{2n-1} f(x_i) + 2 \sum_{i=2, \text{even}}^{2n-2} f(x_i) + f(x_{2n}) \right] \]

Step 5: Explicitly write the formula

\[ \int_a^b f(x) \, dx \approx \frac{b - a}{6n} \left[ f(x_0) + f(x_{2n}) + 4 \left( f(x_1) + f(x_3) + \cdots + f(x_{2n-1}) \right) + 2 \left( f(x_2) + f(x_4) + \cdots + f(x_{2n-2}) \right) \right] \]

Summary: This formula gives a very good approximation to the integral by approximating the function with quadratic polynomials on pairs of subintervals.

Therefore, the correct option is option 3:

\[ \frac{b - a}{6n} \left[ f(x_0) + f(x_{2n}) + 4 \sum_{\text{odd } i=1}^{2n-1} f(x_i) + 2 \sum_{\text{even } i=2}^{2n-2} f(x_i) \right] \]