The Trapezoidal rule formula is given to be:
![\begin{gathered} \int_a^bf(x)dx\approx(\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=(b-a)/(n) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9dc2jtf4za27ti5jb17bc6exhcuiyfdulo.png)
The question gives:
![\begin{gathered} f(x)=\ln(x^2+9) \\ a=3 \\ b=4 \\ n=3 \\ \therefore \\ \triangle x=(1)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4lkzirvajmqvpkbrawccounfjvujsdkz17.png)
Therefore, divide the interval into n = 3 subintervals of length 1/3 with the following endpoints:
![a=3,(10)/(3),(11)/(3),4](https://img.qammunity.org/2023/formulas/mathematics/college/jc2kayygg7f0ihgr62o2gxziae68lqdv7e.png)
Evaluate the function at the endpoints:
![\begin{gathered} f(x_0)=f(3)=2.89 \\ 2f(x_1)=2f((10)/(3))=6.00 \\ 2f(x_2)=2f((11)/(3))=6.22 \\ f(x_3)=f(4)=3.22 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ac67vvtt1dnesz4bu5yabajr4p0gn0yed1.png)
Sum up the calculated values and multiply by Δx/2:
![\Rightarrow(1)/(3*2)(2.89+6.00+6.22+3.22)=3.06](https://img.qammunity.org/2023/formulas/mathematics/college/3fey2efaqui2i3hxtg7zakiugob9gd9ofo.png)
Therefore, the answer will be:
![\int_3^4\ln(x^2+9)dx\approx3.06](https://img.qammunity.org/2023/formulas/mathematics/college/bptt35y2ytmyoahxu9s7ob7x1662nk6fux.png)