Approximating this integral

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In summary, if g is a continuous function that maps a closed bounded subset of R to R^n, then the definite integral from a to b of g(t) can be approximated by the sum from 0 to n-1 of g(t) (b-a)/n. This is because as n approaches infinity, the sum converges to the integral. To go from the integral to the sum, there are steps that need to be taken and it is not valid to simply claim they are equal. If this approximation is not valid, there is a finite sum that can be used instead. To rigorously show that the approximation sum converges to the integral, it is necessary to use Riemann sums. By taking the sum of
  • #1
e12514
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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?
 
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  • #2
Isn't that almost a Riemann sum?

(if instead of taking [itex]\sum g(t)\frac{b-a}{n}[/itex] you take [itex]\sum g(\xi_n)\frac{b-a}{n}[/itex] where [itex]\xi_n[/itex] is in each interval of the partition, then, in the limit, is exactly the Riemann sum.)
 
  • #3


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+
 

What is an integral?

An integral is a mathematical concept that represents the accumulation of a function over a given interval. It can also be thought of as the area under the curve of a function.

How do you approximate an integral?

To approximate an integral, you can use numerical methods such as the trapezoidal rule, Simpson's rule, or the midpoint rule. These methods involve dividing the interval into smaller subintervals and using the values of the function at specific points within each subinterval to approximate the integral.

Why is it important to approximate integrals?

Approximating integrals is important in many scientific and engineering fields, as it allows us to find estimates for quantities such as area, volume, and average values. It also helps us solve differential equations, which are used to model real-world phenomena.

What are some limitations of approximating integrals?

One limitation of approximating integrals is that the accuracy of the approximation depends on the number of subintervals used. Using too few subintervals can result in a less accurate estimate, while using too many can be computationally expensive. Additionally, these methods may not work well for certain types of functions, such as those with discontinuities or rapidly changing values.

Can you use technology to approximate integrals?

Yes, there are many software programs and calculators that have built-in functions for approximating integrals. These tools often use more advanced numerical methods and can handle a wider range of functions compared to manual methods. However, it's important to understand the underlying concepts and limitations of these methods before relying solely on technology.

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