Question

Evaluate the integral \(I = \int_0^1 \frac{1}{1+x} dx\) by using Simpson's one-third rule with \(h = 0.5\)

• A. 0.7945
• B. 0.5945
• C. 0.6945
• D. 0.4945

Hint:

Simpson's one-third rule is accurate for polynomials up to degree 3.
Use \(n\) even and \(h = (b-a)/n\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Simpson's one-third rule is a numerical integration method.
It requires an even number of subintervals.

Step 2: Key Formula or Approach:
For \(n = 2\) (since \(h = 0.5\) from 0 to 1),
\[ \int_a^b f(x) dx \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + f(x_2) \right]. \]
Here, \(x_0 = 0\), \(x_1 = 0.5\), \(x_2 = 1\).

Step 3: Detailed Explanation:
Given \(f(x) = \frac{1}{1+x}\).
Compute: \[ f(0) = 1, \quad f(0.5) = \frac{1}{1.5} = \frac{2}{3} \approx 0.6667, \quad f(1) = \frac{1}{2} = 0.5. \]
Apply Simpson's rule: \[ I \approx \frac{0.5}{3} \left[ 1 + 4 \times \frac{2}{3} + 0.5 \right] = \frac{0.5}{3} \left[ 1 + \frac{8}{3} + 0.5 \right]. \]
Simplify inside: \[ 1 + 2.6667 + 0.5 = 4.1667. \]
Multiply: \[ I \approx \frac{0.5}{3} \times 4.1667 = \frac{2.08335}{3} = 0.69445. \]
Thus, \(I \approx 0.6945\), which matches option (C).
The exact value is \(\ln 2 \approx 0.6931\), so the approximation is close.