Aprroximating the area of a surface of revolution using simpsons rule

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Consider the surface of revolution generated from rotating the curve $3xe^x$ from $0 \leq x \leq 1$ about the $x$-axis.

So, the integral will look like:

$$\int_0^1 6 \pi e^xx \sqrt{(3e^xx+3e^x)^2+1} \, dx$$

I need to approximate this integral using simpsons rule with $n=10$... After a billion attempts I can't seem to get it right! The correct answer is supposed to be 209.894506

Can somebody walk me through the steps on this one? I feel like i'm going crazy

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If $n=10$ then $\Delta x = (1-0)/10 = 0.1$. Now $$ \begin{split} \int_a^b f(x)\ dx \approx \frac{\Delta x}{3} \left[f(0) + 4f(0.1) + 2f(0.2) \\ + 4f(0.3) + 2f(0.4) \\ + 4f(0.5) + 2f(0.6) \\ + 4f(0.7) + 2f(0.8) \\ + 4f(0.9) + f(1.0)\right] \end{split} $$