Answer :
To find the value of the integral [tex]\int_0^1 f(x) \, dx[/tex] using the Trapezoidal rule, we will approximate the area under the curve using trapezoids. Given are the values:
[tex]f(0) = 5, \quad f(0.5) = 6, \quad f(1) = 7.[/tex]
Step-by-step, the procedure includes:
Divide the interval [0,1] into sub-intervals:
In this case, since we have values given at [tex]x = 0, 0.5, \text{and} 1[/tex], we have two sub-intervals: [tex][0, 0.5][/tex] and [tex][0.5, 1][/tex].
Apply the Trapezoidal rule:
The Trapezoidal rule formula for an interval [tex][a, b][/tex] is:
[tex]\int_a^b f(x) \, dx \approx \frac{b-a}{2} \left( f(a) + f(b) \right).[/tex]
We will apply this formula over the two sub-intervals:
For the interval [tex][0, 0.5][/tex]:
[tex]\int_0^{0.5} f(x) \, dx \approx \frac{0.5 - 0}{2} \left( f(0) + f(0.5) \right) = 0.25 \times (5 + 6) = 2.75.[/tex]For the interval [tex][0.5, 1][/tex]:
[tex]\int_{0.5}^{1} f(x) \, dx \approx \frac{1 - 0.5}{2} \left( f(0.5) + f(1) \right) = 0.25 \times (6 + 7) = 3.25.[/tex]
Add the areas of the trapezoids to estimate the entire integral over [tex][0, 1][/tex]:
[tex]\int_0^1 f(x) \, dx = 2.75 + 3.25 = 6.[/tex]
Therefore, the value of the integral [tex]\int_0^1 f(x) \ dx[/tex] using the Trapezoidal rule is [tex]6[/tex].
This approach is intuitive because it approximates the function [tex]f(x)[/tex] by linear segments between the given points and calculates the area under these linear segments, which is a good approximation if the function is reasonably smooth over the interval.
