From 95f378fa055e56d64128a8d51bc637a72ebb0840 Mon Sep 17 00:00:00 2001
From: arc <zleyyij@users.noreply.github.com>
Date: Tue, 15 Apr 2025 09:56:58 -0600
Subject: [PATCH] vault backup: 2025-04-15 09:56:58

---
 education/math/MATH1210 (calc 1)/Integrals.md | 8 +++++++-
 1 file changed, 7 insertions(+), 1 deletion(-)

diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md
index fc0a915..4a54007 100644
--- a/education/math/MATH1210 (calc 1)/Integrals.md	
+++ b/education/math/MATH1210 (calc 1)/Integrals.md	
@@ -178,4 +178,10 @@ $$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$
 13. $= \frac 1 2 \int_{1/2}^5 (x^2 - x^{-2})dx$: Find the indefinite integral
 14. $= \dfrac{1}{2} (\frac{1}{3}x^3 - x^-1)\Big|_{1/2}^5$ : Plug and chug
 15. $= (\frac{125}{6} - \frac{1}{10}) - (\frac{1}{48} - 1)$ 
-16. $=(\frac{5000}{240} - \frac{24}{240}) - (\frac{5}{240} - \frac{240}{240})$ 
\ No newline at end of file
+16. $=(\frac{5000}{240} - \frac{24}{240}) - (\frac{5}{240} - \frac{240}{240})$ 
+
+> Find the length of the curve $y = \sqrt{1 - x^2}$
+1. $y$ has a domain of $[-1, 1]$
+2. $y' = \dfrac{1}{2}(1-x^2)^{-1/2}(-2x)$
+3. $= -\dfrac{x}{\sqrt{1 - x^2}}$
+4. $L = \int_{-1}^1 \sqrt{1 + (-\frac{x}{\sqrt{1-x^2}})}$ 
\ No newline at end of file