Question

Riemann sum integration problem:

How do I integrate INTEGRAL of sqrt(9 - x^2) dx from 0 to 3?

Basically, I should get R_n and I have to take the limit of R_n as n approaches infinity, but what is R_n? The root has me stuck.

Any input would be greatly appreciated!

Answer 1

The precise steps you take depend on what kind of Riemann sum you choose to use. If you'd like to use inscribed rectangles, for example, you have
\[\int_0^3\sqrt{9-x^2}=\sum_{i=1}^nf(x^*_i)\Delta x\]
where \(x^*_i\) is the value in the interval \((x_i,x_{i+1})\) that takes on the minimal value of \(f(x)\). (Since \(\sqrt{9-x^2}\) is decreasing on \([0,3]\), you'd use \(x_i^*=x_{i+1}\)).

Whatever you choose, as \(n\to\infty\), \(x_i^*\) is essentially the same, but the expression you'll be working with will be different.

Answer 2

Slight correction: the starting index for the series need not necessarily be \(i=1\), nor does the upper index have to be \(n\); another nuance that depends on the summation approach.

I should also mention that using inscribed rectangles doesn't always imply \(x_i^*=x_{i+1}\) for a decreasing function like this. I meant to say that this applies if you're using an absolute lower approximation, i.e. rectangles inscribed by the curve based on their rightmost endpoints.

I'll go ahead and use circumscribed rectangles to keep the notation simpler, and that these rectangles will be assigned according to the absolute upper approximation, so that \(x_i^*=x_i\).
\[\int_0^3\sqrt{9-x^2}~dx=\sum_{i=0}^{n-1}f(x_i)\Delta x\]
where \(\Delta x=x_{i+1}-x_i\).

Answer 3

Another slight correction:
\[\int_0^3\sqrt{9-x^2}~dx=\color{red}{\lim_{n\to\infty}}\sum_{i=0}^{n-1}f(x_i)\Delta x\]or
\[\int_0^3\sqrt{9-x^2}~dx\color{red}\approx\sum_{i=0}^{n-1}f(x_i)\Delta x\]

Answer 4

But, the question I'm trying to do is asking me to compute the limit (and not just set it up).

How would I compute that definite integral using right endpoints of rectangles?

I can't seem to get rid of the root so that I could use the Σ_(i=0)^n i = n(n+1)/2 and Σ_(i=0)^n i^2 = n(n+1)(2n+1)/6 formulas.