Question

The value of the integral \( \int_1^3 \frac{2}{x} \, dx \), when evaluated by using Simpson’s \( \frac{1}{3} \) rule on two equal subintervals each of length 1, is ...........

• A. 2.00
• B. 2.24
• C. 2.19
• D. 2.22

Hint:

Use Simpson’s rule when the interval is divided into an even number of subintervals, and apply the weights 1-4-1 accordingly for three points.

Answer 1

The Correct Option is D

We are given the integral: \[ \int_1^3 \frac{2}{x} \, dx \]
Simpson's \( \frac{1}{3} \) rule on 2 subintervals of equal width \( h = 1 \):
Let \( a = 1, b = 3 \Rightarrow h = \frac{3 - 1}{2} = 1 \)
Points: \( x_0 = 1, x_1 = 2, x_2 = 3 \)
Now evaluate the function:
\( f(x) = \frac{2}{x} \)
\[ f(x_0) = f(1) = 2, f(x_1) = f(2) = 1, f(x_2) = f(3) = \frac{2}{3} \]
Simpson's rule formula:
\[ \int_1^3 \frac{2}{x} dx \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + f(x_2) \right] = \frac{1}{3}(2 + 4 \cdot 1 + \frac{2}{3}) = \frac{1}{3} \cdot \left(6 + \frac{2}{3}\right) = \frac{1}{3} \cdot \frac{20}{3} = \frac{20}{9} \approx 2.22 \]