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?
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$\begingroup$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
$\endgroup$ $\begingroup$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
$$\lim_{n\to\infty}\left[\sum_{k=1}^n \left(\frac{3}{n}+\frac{9}{n^2}\sqrt{2kn-k^2}\right)\right]$$
$\endgroup$ $\begingroup$First off, I don't think your expression of the defining Riemann sums is correct, so you may want to re-evaluate how you arrived at it.
In any event, you aren't going to be able to evaluate this integral using Riemann sums so easily. Using the right-hand points for the evaluation of our Riemann sum, we have $$\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left(1+\sqrt{9-\left(-3+\frac{3i}{n}\right)^{2}}\right)=\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^{n}\left(1+\sqrt{9-\left(9-\frac{18i}{n}+\frac{9i^{2}}{n^{2}}\right)}\right)=\lim_{n\to\infty}\left(\frac{3}{n}\sum_{i=1}^{n}1+\frac{3}{n}\sum_{i=1}^{n}\sqrt{\frac{18i}{n}-\frac{9i^{2}}{n^{2}}}\right)=3+\lim_{n\to\infty}\frac{9}{n^{2}}\sum_{i=1}^{n}\sqrt{2in-i^{2}}$$
The expression $\sum_{i=1}^{n}\sqrt{2in-i^{2}}$ is not easily simplified in a manner such that the limit can be explicitly evaluated. Thus, if you want the exact answer, then you need to apply the usual integration techniques. The first integral is just $$\int_{-3}^{0}\;dx=3.$$
Since $9-(3\sin\theta)^{2}=9\cos^{2}\theta,$ we are led to the substituion $x=3\sin\theta$ and so $dx=3\cos\theta\;d\theta.$ The inverse of this transformation is $\theta=\sin^{-1}(x/3)$. This gives us for the second integral $$\int_{-3}^{1}\sqrt{9-x^{2}}\;dx=9\int_{\sin^{-1}(-1)}^{\sin^{-1}(0)}|\cos\theta|\cos\theta\;d\theta=9\int_{-\pi/2}^{0}\cos^{2}\theta\;d\theta=\frac{9}{2}\Bigg[x+\sin x\cos x\Bigg]_{x=-\pi/2}^{x=0}=\frac{9\pi}{4}.$$
The answer is therefore $$\frac{12+9\pi}{4}.$$
Alternatively, you can observe that the second integral is the just the area of a quarter circle of radius 3, so is seen to be $9\pi/4$. The first is even simpler, a rectangle of dimensions 3 by 1, so is equal to $3$.
$\endgroup$ $\begingroup$you shouldn't try to use a Riemann sum for this. Your instructor definitely isn't expecting you to solve that way. This is a "net area" problem, check your textbook for "net area".
The short version is this: ⺆r²=the area of a circle. In this case, r will be the same as x. So basically, you know x has to be 3 by setting the equation of the circle equal to zero, which is how you always solve for x, right? This is true for all half circle equations of this format. So you know you have a radius of 3 now. You also know that the location of your interval [-3,0] is such that your interval will cut the circle into quarters >> look at the graph on your calculator. Yes, you should always graph every line you're ever given on your calculator, lol. You can see a half circle slightly raised off the x-axis, right? So from -3 to 0 on the x-axis, you only have a quarter of the total possible circle there. so you need the area of 1/4 of your circle, since only a fourth of the total possible circle lies within your interval. so (1/4)Area of your circle = (1/4)⺆r², where r=3. This is true since the area of the total possible circle would just be ⺆r², where r=3, like we discussed. This gives you (1/4)9⺆, which multiplies out to (9/4)⺆. That's the area of a quarter of the total circle, half of which you see on your graphing calculator, with only that quarter we calculated the area for being in the necessary interval of [-3,0].
Now, you need to remember that there's space under your circle thats just open, like we were saying. Like, the edges of your circle don't touch the x-axis. yeah, you have to find the area of that also, and add it to the area of the circle part that you already found. Net area of an interval is ALWAYS from the line you see, no matter what shape it is, all the way to the x-axis. Just quickly sketch the graph and color in the entire area from the line (in this case, that half circle) down to the x-axis, and that's the total area you need to find. Don't just stop at the edges of the circle. So that shape under your circle is actually just an invisible rectangle, isn't it? Sketch that rectangle onto your hand drawn graph if necessary, so you can see it. You need to add the area of that quarter circle, with the area of the invisible rectangle underneath the circle. So the area of a rectangle is base times height, and in our case, it should be 1 x 3 = 3. So the area of the quarter circle is (9/4)⺆ and the area of the invisible rectangle underneath that quarter circle is 3. So add them together to get: (9/4)⺆ + 3 = net area.
Yes, net area of this format is that easy. And leave your answer in the above form, don't do anything else. You dont need to, because ⺆ isn't a variable, remember? Its a constant number, so you have all constant numbers in your answer, and this form is the exact answer. That's what you want. As for a more formal looking solution, just write out:
Net area = (base x height) + (1/4)⺆r² as the formula you created to solve and plug in your values.
So DONT try a Riemann Sum, you don't have to! It's completely unnecessary, and your instructor won't accept a Riemann Sum as a correct answer to a net area problem. Your answer should always be constant numbers when you're solving for net area. I hope this helps!
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