Evaluate the integral by interpreting it in terms of areas. integral: ∫−30(1+9−x2)dx

Step 1
First we break the integral into two parts:
${\displaystyle {\int }_{-3}^0(1+\sqrt{9-x^2})dx={\int }_{-3}^{0}1dx+{\int }_{-3}^0\sqrt{9-x^2}dx}$
The first term ${\displaystyle {\int }_{-3}^{0}1dx}$ represents area of rectangle with base ${\displaystyle b=3}$ and height ${\displaystyle h=1}$
${\displaystyle {\int }_{-3}^{0}1dx=b\times h=3}$
${\displaystyle {\int }_{-3}^0\sqrt{9-x^2}dx}$ represent area of quarter circle with radius ${\displaystyle r=3}$
${\displaystyle {\int }_{-3}^0\sqrt{9-x^2}dx=\frac{1}{4}\pi \times 3^2=\frac{9\pi }{4}}$
${\displaystyle {\int }_{-3}^0(1+\sqrt{9-x^2})dx={\int }_{-3}^{0}1dx+{\int }_{-3}^0\sqrt{9-x^2}dx}$
${\displaystyle {\int }_{-3}^0(1+\sqrt{9-x^2})dx=3+\frac{9\pi }{4}}$

Step 1
The given integral is
${\displaystyle I={\int }_{-3}^0(1+\sqrt{9-x^x})dx}$
${\displaystyle I={\int }_{-3}^{0}1\times x+{\int }_{-3}^0\sqrt{9-x^2}dx}$
${\displaystyle I=A_1+A_2}$
we can plot area A1 as follows

we can plot area A2 as follows

Hence we can calculate the integral as
${\displaystyle I=W\times H+\frac{\pi r^2}{4}}$
${\displaystyle I=3\times 1+\frac{\pi {\left(3\right)}^2}{4}}$
${\displaystyle I=3+\frac{9\pi }{4}}$

The expression you are having trouble with seems to be incorrect.
You have
$f(x)=1+\sqrt{9-x^2}$
Since
$a=-3$ and $b=0$
you have
$\mathrm{\Delta }x=\frac{3}{n}and{x}_k=-3+\frac{3k}{n}$
Applying your definition:
$f(x_k)=1+{\sqrt{9-(-3+\frac{3k}{n})}}^2$
$=1+\sqrt{18}\frac{k}{n}-9\frac{k^2}{n^2}$
$=1+\frac{3}{n}\sqrt{2kn-k^2}$
From this, we see that
$f(x_k)\mathrm{\Delta }x=\frac{3}{n}(1+\frac{3}{n}\sqrt{2kn-k^2})$
$=\frac{3}{n}+\frac{9}{n^2}\sqrt{2kn-k^2}$
Then the final expression should be
$\underset{n\Rightarrow \mathrm{\infty }}{lim}[\sum _{k=1}^n(\frac{3}{n}+\frac{9}{n^2}\sqrt{2kn-k^2})]$