let f(0)=1, f(1/2)=2, f(1)=3, and f(3/2)=4. which of the following is the best approximation for ∫₀²f(x) dx?

Explanation:

Step1: Use trapezoidal - rule formula

The trapezoidal - rule for approximating $\int_{a}^{b}f(x)dx$ with $n$ sub - intervals is $T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots + 2f(x_{n - 1})+f(x_n)]$, where $\Delta x=\frac{b - a}{n}$ and $x_i=a + i\Delta x$. Here, $a = 0$, $b = 2$, and we will use $n = 4$ sub - intervals with $\Delta x=\frac{2-0}{4}=0.5$. The points are $x_0 = 0$, $x_1=\frac{1}{2}$, $x_2 = 1$, $x_3=\frac{3}{2}$.

Step2: Substitute values into the formula

$T_4=\frac{0.5}{2}[f(0)+2f(\frac{1}{2})+2f(1)+2f(\frac{3}{2})]$. Substitute $f(0)=1$, $f(\frac{1}{2}) = 2$, $f(1)=3$, and $f(\frac{3}{2}) = 4$ into the formula: $T_4=\frac{0.5}{2}[1+2\times2 + 2\times3+2\times4]$. First, calculate the values inside the brackets: $1+4 + 6+8=19$. Then, $T_4=\frac{0.5}{2}\times19=\frac{19}{4}=4.75$.

Answer:

$4.75$