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

x	0	0.25	0.5	0.75	1.0
f(x)	0.9	2.0	1.5	1.8	0.4

Using Simpson’s one-third rule,

${\int }_0^{1}f\left(x\right)dx\approx \text{\_\_\_\_\_}\text{ (rounded off to 2 decimal place)}$

The correct answer is 1.63.

To approximate the integral ${\int }_0^{1}f\left(x\right)dx$ using Simpson's one-third rule, we first analyze the given data points from the table:

x₀ = 0, f(x₀) = 0.9
x₁ = 0.25, f(x₁) = 2.0
x₂ = 0.5, f(x₂) = 1.5
x₃ = 0.75, f(x₃) = 1.8
x₄ = 1.0, f(x₄) = 0.4

The interval size h between successive points is constant:
h = 0.25 - 0 = 0.5 - 0.25 = 0.25

The number of subintervals is n = 4, which is even, allowing us to apply Simpson's one-third rule.

Simpson's one-third rule formula for n = 4 is given by:

${\int }_a^{b}f\left(x\right)dx\approx \frac{h}{3}\left[f\right(x_0)+f(x_4)+4(f\left(x_1\right)+f\left(x_3\right))+2f(x_2\left)\right]$

Substituting the values from the table into the formula, we get:

${\int }_0^{1}f\left(x\right)dx\approx \frac{0.25}{3}[0.9+0.4+4(2.0+1.8)+2(1.5\left)\right]$

Now, we compute the expression inside the brackets:

First and last terms: 0.9 + 0.4 = 1.3
Odd terms multiplied by 4: 4 * (2.0 + 1.8) = 4 * 3.8 = 15.2
Even term multiplied by 2: 2 * 1.5 = 3.0

Summing these values together:

1.3 + 15.2 + 3.0 = 19.5

Finally, we multiply by h/3:

$\mathrm{Integral}\approx \frac{0.25}{3}\times 19.5=\frac{4.875}{3}=1.625$

Rounding 1.625 off to 2 decimal places gives 1.63.