Can Riemann Sums Approximate Integrals of Vector-Valued Functions?

[e12514]

If g maps a closed bounded subset of R to R^n
g : [a,b] -> R^n
and g is continuous,

can the definite integral
(integral from a to b) g(t) dt

be approximated by
(SUM from 0 to n-1 ) ( g(t) (b-a)/n ) ?
(because as taking n->oo gives the integral?)

If so, what are the steps needed to go from the integral to the sum (obviously it's not valid just to claim they're eqaul)?
Or if not, what's the finite sum that approximates that integral?

And also, how do we actually rigourously show that the approximation sum converges to a limit equal to the integral?

 

[AiRAVATA]

Isn't that almost a Riemann sum?

(if instead of taking $\sum g(t)\frac{b-a}{n}$ you take $\sum g(\xi_n)\frac{b-a}{n}$ where $\xi_n$ is in each interval of the partition, then, in the limit, is exactly the Riemann sum.)

 

[tacman]

Yes, the definite integral can be approximated by the sum given in the question. This is known as the Riemann sum, which is a well-known method for approximating integrals. The steps to go from the integral to the sum are as follows:

1. Divide the interval [a,b] into n equal subintervals. This can be done by choosing n points a = t_0 < t_1 < ... < t_n = b such that t_i - t_{i-1} = (b-a)/n for i = 1,2,...,n.

2. Approximate the integral on each subinterval [t_{i-1}, t_i] by the value of g at a point in that interval. This can be done by choosing a point c_i in the interval and using g(c_i) as the approximation.

3. Multiply each approximation by the length of the corresponding subinterval (b-a)/n. This gives us the Riemann sum: SUM from i=1 to n (g(c_i) (b-a)/n).

To show that this approximation sum converges to the limit equal to the integral, we need to use the definition of a Riemann integral. This involves taking the limit of the Riemann sums as n approaches infinity. If this limit exists and is equal to the integral, then we have shown that the Riemann sum converges to the integral.

In order to prove this rigorously, we need to use the concept of continuity of g. Since g is continuous on the closed bounded interval [a,b], it is also uniformly continuous. This means that for any given epsilon > 0, there exists a delta > 0 such that for any x,y in [a,b], if |x-y| < delta, then |g(x) - g(y)| < epsilon.

Using this property, we can show that the Riemann sum converges to the integral. Let M be the maximum value of g on the interval [a,b]. Then, for any c_i chosen in the subinterval [t_{i-1},t_i], we have |g(c_i) - g(x)| < epsilon whenever |c_i - x| < delta. This means that |g(c_i)| < M + epsilon for all i. Therefore, the Riemann sum is bounded above by n(M + epsilon)(b-a)/n = (M+