Answer:
The integral is 0.5090
Explanation:
Given:
[tex]\begin{gathered} f(x)=\frac{1}{(x-1)^2} \\ \\ [a,b]=[2,3] \\ n=4 \end{gathered}[/tex]
We have:
[tex]\Delta x=\frac{b-a}{4}=\frac{3-2}{4}=\frac{1}{4}[/tex]
The starting point is a = 2 and end point is b = 3. Then
[tex]\begin{gathered} x_0=2 \\ \\ x_1=2+\frac{1}{4}=\frac{9}{4} \\ \\ x_2=\frac{9}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2} \\ \\ x_3=\frac{11}{4} \\ \\ x_4=\frac{12}{4}=3 \end{gathered}[/tex]
Finally, we have the integral as:
[tex]\begin{gathered} I=\frac{\Delta}{2}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)] \\ \\ =\frac{1}{8}[f(2)+2f(\frac{9}{4})+2(\frac{5}{2})+2f(\frac{11}{4})+f(3)] \\ \\ =\frac{1}{8}[1+\frac{32}{25}+\frac{8}{9}+\frac{32}{49}+\frac{1}{4}] \\ \\ =0.5090 \end{gathered}[/tex]
