From 6d430c98fa7f8dff0d599328e0eca340b3bb1da2 Mon Sep 17 00:00:00 2001 From: arc Date: Tue, 15 Apr 2025 09:31:58 -0600 Subject: [PATCH] vault backup: 2025-04-15 09:31:58 --- education/math/MATH1210 (calc 1)/Integrals.md | 9 ++++++++- 1 file changed, 8 insertions(+), 1 deletion(-) diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md index b9c90cd..3b866ef 100644 --- a/education/math/MATH1210 (calc 1)/Integrals.md +++ b/education/math/MATH1210 (calc 1)/Integrals.md @@ -157,4 +157,11 @@ $$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$ 1. $L = \int_{-1}^8 \sqrt{1 + (-\frac{5}{12})^2} dx$ 2. $= \int_{-1}^8 \sqrt{1 + \frac{25}{144}} dx$ -3. = $\int_{-1}^8 \sqrt{\frac{169}{144}} \ No newline at end of file +3. = $\int_{-1}^8 \sqrt{\frac{169}{144}}dx$ +4. $= \int_{-1}^8 \frac{13}{12} dx$ +5. $\frac{13}{12} x \Big| _{-1}^8$ + +> Find the distance from the point ${\frac{1}{2}, \frac{49}{48}}$ to the point $(5, \frac{314}{15})$ along the curve $y = \dfrac{x^4 - 3}{6x}$ +1. $y' = \dfrac{4x^3(6x) - (x^4 + 3)6}{36x^2}$ +2. $= \dfrac{24x^4 -6x^4 - 18}{36x^2}$ +3. \ No newline at end of file