vault backup: 2025-04-15 10:01:58

.obsidian/plugins/obsidian-git/data.json

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

@@ -178,4 +178,15 @@ $$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$
 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. $=(\frac{5000}{240} - \frac{24}{240}) - (\frac{5}{240} - \frac{240}{240})$ 
+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. $L = \int_{-1}^1 \sqrt{1 + (-\dfrac{x}{\sqrt{1-x^2}})^2}dx$ 
+5. $L = \int_{-1}^1 \sqrt{1 + \dfrac{x^2}{1-x^2}}dx$ 
+6. $L = \int_{-1}^1 \sqrt{\dfrac{1}{1-x^2}}dx$ 
+7. $L = \int_{-1}^1 \dfrac{1}{\sqrt{1-x^2}}dx$ 
+8. $L = \arcsin(x) \Big|_{-1}^1$ 
+9. $\arcsin($