From a8fbf79e7739b6621d42a0328153e3d362197a78 Mon Sep 17 00:00:00 2001 From: arc Date: Tue, 15 Apr 2025 09:16:58 -0600 Subject: [PATCH] vault backup: 2025-04-15 09:16:58 --- education/math/MATH1210 (calc 1)/Integrals.md | 10 +++++++++- 1 file changed, 9 insertions(+), 1 deletion(-) diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md index 6e7c1c1..0c5328a 100644 --- a/education/math/MATH1210 (calc 1)/Integrals.md +++ b/education/math/MATH1210 (calc 1)/Integrals.md @@ -135,4 +135,12 @@ With a series of points on a curve we can find the distance between each point. As we increase $n$, the precision of which the curve is estimated increases. This means that: -$$ len(curve) = \lim_{n \to \infty} \sum_{i=1}^{n}(\p{}$$ \ No newline at end of file +$$ \text{length of a curve} = \lim_{n \to \infty} \sum_{i=1}^{n}(\text{length of line segment)}$$ +Using the distance formula, we know that the length of the line segment can be found with: +$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ +1. So the entire equation is: +$$ \text{length of a curve} = \lim_{n \to \infty} \sum_{i=1}^{n}(\sqrt{(x_i - x_{i-1})^2 + (y_i - y_{i-1})^2}) $$ + This can also be described as: +$$ \text{length of a curve} = \lim_{n \to \infty} \sum_{i=1}^{n}(\sqrt{(\Delta x)^2 +(\Delta y)^2}) $$ +2. Using the mean value theorem: +$$ \lim_{n \to \infty} \sum_{i = 1}^n) $$ \ No newline at end of file