Question

The values of a function \( f \) obtained for different values of \( x \) are shown in the table below.

Using Simpson’s one-third rule, approximate the integral \[ \int_0^1 f(x) \, dx \quad \text{(rounded off to 2 decimal places)}. \]

Hint:

Use Simpson's one-third rule for functions with evenly spaced intervals, and ensure the correct application of weights (4 for odd indices, 2 for even indices).

Answer 1

We use Simpson's one-third rule to approximate the integral: \[ \int_0^1 f(x) \, dx \approx \frac{h}{3} \left[ f(x_0) + 4 \sum_{i=1, \, {odd}} f(x_i) + 2 \sum_{i=2, \, {even}} f(x_i) + f(x_n) \right] \] Where \( h \) is the step size, calculated as: \[ h = \frac{b - a}{n} = \frac{1 - 0}{4} = 0.25 \] Using the table values for \( f(x) \): \[ \int_0^1 f(x) \, dx \approx \frac{0.25}{3} \left[ 0.9 + 4(2.0 + 1.8) + 2(1.5) + 0.4 \right] \] Calculating: \[ \int_0^1 f(x) \, dx \approx \frac{0.25}{3} \left[ 0.9 + 4(3.8) + 3.0 + 0.4 \right] \] \[ \int_0^1 f(x) \, dx \approx \frac{0.25}{3} \left[ 0.9 + 15.2 + 3.0 + 0.4 \right] = \frac{0.25}{3} \times 19.5 \] \[ \int_0^1 f(x) \, dx \approx \frac{4.875}{3} \approx 1.625 \] Thus, the integral is approximately \( \boxed{1.63} \).