/sci/ - Science & Math

>>10307818
(continued)
If you try to approximate it with a Riemann sum, you get for each subinterval a product of the form f(c[i])*(x[i] - x[i - 1]) and you sum all the subintervals to get the Riemann sum. (Remember that c(i) is a random point in the interval [x(i - 1), x(i)].) Each term is a product where a random value of the function f in the interval [x(i - 1), x(i)] (the value f(c[i])) is multiplied by the length of interval (namely, (x[i] - x[i - 1])). In other words, each term of the Riemann sum is a rectangle with base (x[i] - x[i - 1]) and height f(c[i]). If the interval [x(i - 1), x(i)] is reasonably small, then the value of f will not vary that much during this time interval, and therefore the area under the graph of f can be approximated by taking a random value f(c[i]) in that interval and multiplying it by the length of the interval itself. Summing up all these small rectangles you get the Riemann sum. What does each rectangle represent? If f is the velocity of particle P and the interval [x(i - 1), x(i)] represents a small time interval, then the product f(c[i])*(x[i] - x[i - 1]) (which is the area of the rectangle) represents the distance traveled by particle P while ASSUMING that during that time interval P had a CONSTANT velocity equal to f(c[i]). When you sum all these products, you get a Riemann sum which approximates the distance traveled by the particle over the whole time period [a, b]. You divide the interval [a, b] in many small subintervals so that you can approximate the velocity of the particle during each one of those subintervals with a constant velocity, instead of a variable one. The smaller the intervals, the better the approximation; and as you take the limit of the Riemann sum, you get the exact value of the integral, which tells you exactly how much distance the particle P traveled during the time interval [a, b].

(to be continued)