37. \(\displaystyle \int^0_{-3} (1+\sqrt{9-x^2})*dx\)

the answer is \(\displaystyle 3+\frac{9}{4}*\pi\)

Where .. the heck did \(\displaystyle \pi\) come from?

I was trying to do

\(\displaystyle \frac{b-a}{n}\)

\(\displaystyle = \frac{0-3}{n}\)

\(\displaystyle =\frac{-3}{n}\)

and then \(\displaystyle x_{i}\) as \(\displaystyle \frac{-3i}{n}-3\)

then plugging \(\displaystyle x_{i}\) into \(\displaystyle f(x)*dx\) then solving for the limit. It came out to be a super complicated root problem so I look in the back to see if I'm in the right track.. and theres a \(\displaystyle \pi\) in the answer...

I think I'm missing something here... how is this integral end up having a \(\displaystyle \pi\)?