Comments by "MC116" (@angelmendez-rivera351) on "Definite Integral Using Limit Definition" video.

  1. I should note that, technically, the definition given in the video is not the definition of the integral. If f is a function on [a, b], then you can partition [a, b] into intervals [x(i), x(i + 1)] with x(0) = a and x(n + 1) = b, and you can tag this partition letting t(i) be an element of [x(i), x(i + 1)]. Then the Riemann sum over the tagged partition is the sum of f[t(i)]·[x(i + 1) – x(i)]. The mesh of a partition is given by max[x(i + 1) – x(i)]. The integral is equal to the limit as max[x(i + 1) – x(i)] —> 0 of the Riemann sums. Here, we have that f(x) = 4·x^2 is a function on [1, 4]. So the Riemann sums are the sums of 4·t(i)^2·[x(i + 1) – x(i)]. Now, we have that max[x(i + 1) – x(i)] —> 0, so it is quite natural to have x(i + 1) – x(i) = max[x(i + 1) – x(i)] – d(i), with d(i) —> 0, so x(j) – x(0) = j·max[x(i + 1) – x(i)] – Sum{0 =< i =< j – 1, d(i)}. Let t(i) = x(i) + s(i) = x(0) + max[x(i + 1) – x(i)]·i + s(i) – Sum{0 =< j =< i – 1, d(j)}, with s(i) —> 0. Hence 4·t(i)^2·[x(i + 1) – x(i)] = 4·[1 + s(i) – Sum{0 =< j =< i – 1, d(j)} + max[x(i + 1) – x(i)]·i]^2·{max[x(i + 1) – x(i)] – d(i)} = 4·{[1 + s(i) – Sum{0 =< j =< i – 1, d(j)}]^2 + 2·[1 + s(i) – Sum{0 =< j =< i – 1, d(j)}]·max[x(i + 1) – x(i)]·i + max[x(i + 1) – x(i)]^2·i^2}·{max[x(i + 1) – x(i)] – d(i)} = 4·max[x(i + 1) – x(i)]·[1 + s(i) – Sum{0 =< j =< i – 1, d(j)}]^2 + 8·max[x(i + 1) – x(i)]^2·[1 + s(i) – Sum{0 =< j =< i – 1, d(j)}]·i + 4·max[x(i + 1) – x(i)]^3·i^2 – 4·[1 + s(i) – Sum{0 =< j =< i – 1, d(j)}]^2·d(i) – 8·max[x(i + 1) – x(i)]·[1 + s(i) – Sum{0 =< j =< i – 1, d(j)}]·d(i)·i – 4·max[x(i + 1) – x(i)]^2·d(i)·i^2. Since max[(i + 1) – x(i)] is asymptotically equivalent to K/n for some real K > 0, one can use repeated applications of Tannery's theorem to conclude the limit as max[x(i + 1) – x(i)] —> 0 of the sums of the above is equal to the same limit of the sums of 4·max[x(i + 1) – x(i)] + 8·max[x(i + 1) – x(i)]^2·i + 4·max[x(i + 1) – x(i)]^3·i^2. The sums are given by 4·(n + 1)·max[x(i + 1) – x(i)] + 8·max[x(i + 1) – x(i)]^2·(n^2 + n)/2 + 4·max[x(i + 1) – x(i)]^3·(n^3/3 + n^2/2 + n/6), and as we have max[x(i + 1) – x(i)] —> 0, we get 4·K + 4·K^2 + 4/3·K^3. For reference, we have, from the fundamental theorem of calculus, that the integral is equal to 4/3·(4^3 – 1^3) = 4·(4^2 + 4·1 + 1^2) = 4·(16 + 4 + 1) = 4·21 = 84. So we know 4/3·K^3 + 4·K^2 + 4·K – 84 = 0. This has only one real solution, K = 3. In fact, the way one would go about computing K is by having K = x(n + 1) – x(0) = b – a = 4 – 1 = 3, but proving that this is the case is tedious and very laborious in itself. This is why we do not use the definition of the integral to compute them, and instead, we prove the minimal integral theorems, which are easier to do than computing any given integral, and then use the theorems instead. Also, if someone is wondering how exactly is this more correct, this is because t(i) is not being taken to simply be equal to x(i) or x(i + 1), but is an arbitrary number in the enclosed interval. Otherwise, s(i) = 0 or s(i) = x(i + 1) – x(i) respectively; and x(i + 1) – x(i) is not constant, hence t(i) is not merely a first-degree polynomial function of i. Otherwise, d(i) = 0.
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