The Trapezoidal rule formula is given to be:
[tex]\begin{gathered} \int_a^bf(x)dx\approx\frac{\triangle x}{2}(f(x_o)+2f(x_1)+2f(x_2)+2f(x_3)+...+2f(x_{n-1})+f(x_n) \\ where \\ \triangle x=\frac{b-a}{n} \end{gathered}[/tex]
The question gives:
[tex]\begin{gathered} f(x)=\ln(x^2+9) \\ a=3 \\ b=4 \\ n=3 \\ \therefore \\ \triangle x=\frac{1}{3} \end{gathered}[/tex]
Therefore, divide the interval into n = 3 subintervals of length 1/3 with the following endpoints:
[tex]a=3,\frac{10}{3},\frac{11}{3},4[/tex]
Evaluate the function at the endpoints:
[tex]\begin{gathered} f(x_0)=f(3)=2.89 \\ 2f(x_1)=2f(\frac{10}{3})=6.00 \\ 2f(x_2)=2f(\frac{11}{3})=6.22 \\ f(x_3)=f(4)=3.22 \end{gathered}[/tex]
Sum up the calculated values and multiply by Δx/2:
[tex]\Rightarrow\frac{1}{3\times2}(2.89+6.00+6.22+3.22)=3.06[/tex]
Therefore, the answer will be:
[tex]\int_3^4\ln(x^2+9)dx\approx3.06[/tex]
