Question

Express the integral as a limit of sums.
int_{0}^{π}sin5xdx

Answer 1

It is just 2/5

Answer 2

int = sum; dx = \(\Delta x\)

Answer 3

maybe?

Answer 4

oh, you mean like approximation; midpoint, simpson..

Answer 5

limit of sums reminds me of reimann stuff

Answer 6

It comes back to the partition question you asked earlier.

The Riemann integral is defined to be the limit of a Riemann sum. Here's such a sum where the interval [0,pi] has just two partitions, [0,pi/2], [pi/2,pi]. I.e., we've broken up the interval of integration into two intervals, each of length pi/2:

\[ S_n = \frac{\pi}{2} \sin(0) + \frac{\pi}{2} \sin(\pi/2) \]

Answer 7

That is in fact such a sum \( S_2 \), 2 for two partitions of the interval [0,pi].

Answer 8

\[\lim_{\Delta x ->0}\ \sum_{i=0}^{\pi} f(x_i)\Delta x_i\]

Answer 9

You can see that it's an approximation for the area, or the integral. It's a bad approximation, but that's to be expected for a partition with such a small number.

Answer 10

If we broke [0,pi] up into 4 partitions of equal length pi/4, the Riemann sum would be
\[ S_4 = \frac{\pi}{4} \sin(0) + \frac{\pi}{4} \sin(\pi/4) + \frac{\pi}{4} \sin(2\pi/4) + \frac{\pi}{4} \sin(3\pi/4) \]

Answer 11

This is a better approximation of the integral. Following?

Answer 12

yes, sir!

Answer 13

So now we can generalize to n partitions of [0,pi], in which case the Riemann sum is
\[ S_n = \sum_{i=0}^{n-1} \sin(0 + i \frac{\pi}{n}) .\frac{\pi}{n} \]

I wrote the zero in the sin just to emphasize that we're beginning at zero.

Answer 14

If now the limit of the Riemann sum S_n exists as n --> infinity, then the Riemann integral over that interval exists and we write
\[ \lim_{n \rightarrow \infty} S_n = \int_0^{\pi} \sin x \ dx \]
The definition of the symbols on the right hand side of the equation is the limit.

Answer 15

(There are some minor technical wrinkles here, but this is substantially correct.)

Answer 16

And for your problem you need to use sin(5x), not sin(x) as I have.

Answer 17

yeah, I got that... Thanks :)