From 8810f2f9cfced2d1ac4c059ab4aa15a5b7b3c305 Mon Sep 17 00:00:00 2001
From: arc <zleyyij@users.noreply.github.com>
Date: Tue, 15 Apr 2025 09:41:58 -0600
Subject: [PATCH] vault backup: 2025-04-15 09:41:58
---
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 40e50d2..255b0e4 100644
--- a/education/math/MATH1210 (calc 1)/Integrals.md
+++ b/education/math/MATH1210 (calc 1)/Integrals.md
@@ -168,4 +168,10 @@ $$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$
4. $= \dfrac{x^4 - 1}{2x^2}$: Factor out $18$ again
5. $L = \int_{1/2}^5 \sqrt{1 + (\dfrac{4x-1}{2x^2})^2}dx$ : Use the length formula
6. $= \int_{1/2}^5 \sqrt{1 + \dfrac{x^8 - 2x^4 + 1}{x^4}} dx$: Apply the $^2$
-7. $= \int_{1/2}^5 \sqrt{\dfrac{4x^4 + x^8 -2x^4 + 1}{4x^4}}$: Set $1 = \dfrac{4x^4}{4x^4}$ and add
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+7. $= \int_{1/2}^5 \sqrt{\dfrac{4x^4 + x^8 -2x^4 + 1}{4x^4}}dx$: Set $1 = \dfrac{4x^4}{4x^4}$ and add
+8. $= \int_{1/2}^5 \sqrt{\dfrac{x^8 + 2x^4}{4x^4}}dx$
+9. $\int_{1/2}^5 \sqrt{\dfrac{(4x+1)^2}{4x^4}}$
+10. = $\int_{1/2}^5 \dfrac{x^4 + 1}{2x^2}dx$
+11. $= \frac{1}{2}\int_{1/2}^5 \dfrac{x^4 + 1}{x^2}$
+12. $= \dfrac{1}{2} \int_{1/2}^5 (x^4 + 1)(x^{-2})dx$
+13. $= \frac 1 2 \int_{1/2}^5 (x^2 - x^$
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