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education/math/MATH1210 (calc 1)/Integrals.md

@@ -119,3 +119,20 @@ This theorem tells us that a continuous function on the closed interval will obt
 - $\int \dfrac{1}{u\sqrt{u^2 - a^2}} du = \dfrac{1}{a}arcsec(\dfrac{|u|}{a}) + C$  
 
 # Length of a Curve
+## Review of the Mean Value Theorem
+If $f$ is a continuous function on the interval $[a, b]$ and differentiable on $(a, b)$, then there exists a number $c$ in the interval $(a, b)$ such that:
+
+$$ f'(c) = \dfrac{f(b) - f(a)}{b - a} $$
+
+This also implies that for some $c$ in the interval $(a, b)$:
+$$ f(b) - f(a) = f'(c)(b-a) $$
+
+## Intuitive Approach
+Given that we divide a curve into $n$ sub-intervals, and we can find the location of the right endpoint of each interval.
+
+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:
+$$ len(curve) = \lim_{n \to \infty} \sum_{i=1}^{n}(\p{}$$