Definite Integral of $1+\sqrt{9-x^2}$? Right, so I'm to find the definite integral (interpreting it as an area)...
$\int^0_{-3}(1+\sqrt{9-x^2})dx$
How do I go about doing this?
I am to specifically use the following theorem to work it out...
If $f$ is integrable on $[a,b]$ then...
$\int^b_a f(x)\,dx = \lim_{n\rightarrow \infty}\sum^n_{i=1}f(x_i)\Delta x$
where $\Delta x = \frac{b-a}{n}$ and $x_i = a + i\Delta x$
I keep getting halfway through and getting stuck with $\lim_{n\rightarrow\infty}\frac{3\sqrt{3}}{n^3}\sum^n_{i=1}\sqrt{4n^2-3i}$
Is this correct? How do I work on from here, if so? The answer given in the book is $3 + \frac{9}{4}\pi$.
Any ideas?

The expression you are having trouble with seems to be incorrect.
You have $\mathrm{f}(x) = 1+\sqrt{9-x^2}$. Since $a=-3$ and $b=0$, you have $\displaystyle{\Delta x = \frac{3}{n}}$ and $\displaystyle{x_k = -3+\frac{3k}{n}}$.
Applying your definition:
\begin{eqnarray*}
\mathrm{f}(x_k) &=& 1+\sqrt{9-\left(-3+\frac{3k}{n}\right)^2} \\ \\ \\
&=&1+\sqrt{18\frac{k}{n}-9\frac{k^2}{n^2}} \\ \\ \\
&=&1+\frac{3}{n}\sqrt{2kn-k^2}
\end{eqnarray*} 
From this, we see that
\begin{eqnarray*}
\mathrm{f}(x_k)\Delta x &=& \frac{3}{n}\left(1+\frac{3}{n}\sqrt{2kn-k^2}\right) \\ \\ \\
&=& \frac{3}{n}+\frac{9}{n^2}\sqrt{2kn-k^2}
\end{eqnarray*}
Then the final expression should be
$$\lim_{n\to\infty}\left[\sum_{k=1}^n \left(\frac{3}{n}+\frac{9}{n^2}\sqrt{2kn-k^2}\right)\right]$$
