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?
Best Answer
This doesn't use your theorem, but it solves it in what I consider to be a fast and efficient way. It doesn't really even use calculus! If you were to graph $y = 1+\sqrt{9-x^2}$, you would see that it is in the shape of a half circle from $(-3,3)$, and then the rest of the graph is simply 1, because the square root is imaginary. From this, it is fairly easy to see that $\int_{-3}^{0}(1 + \sqrt{9-x^2})$ is just a quarter of a circle with radius $3$, with $3$ added, because the circle is elevated one unit over a distance of $3$, and $3 \cdot 1 = 3$. A circle with radius $3$ has an area of $9 \pi$, so a quarter of that circle is $\frac{9}{4}\pi$, which is of course $\frac{9}{4}\pi +3$ when the three is added. If any of this is confusing, then try to graph the function and you'll see what I mean. I'm not sure if you wanted any way to solve the integral, or for us to simplify $\lim_{n \to \infty} \frac{3\sqrt{3}}{n^3} \sum _{i = 1}^{n}(\sqrt{4n^2-3i})$, but this is definitely a fairly quick method of solving the integral itself.
Note: This isn't a complete answer, because it doesn't use the stated theorem, but it was too long for a comment, and it seemed worth mentioning