From 92ae2007944a7df9c6556a8fbd092abc1e7b2570 Mon Sep 17 00:00:00 2001
From: arc <zleyyij@users.noreply.github.com>
Date: Tue, 15 Apr 2025 09:51:58 -0600
Subject: [PATCH] vault backup: 2025-04-15 09:51:58
---
education/math/MATH1210 (calc 1)/Integrals.md | 17 +++++++++--------
1 file changed, 9 insertions(+), 8 deletions(-)
diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md
index a2daa8d..fc0a915 100644
--- a/education/math/MATH1210 (calc 1)/Integrals.md
+++ b/education/math/MATH1210 (calc 1)/Integrals.md
@@ -161,7 +161,8 @@ $$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$
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 $y = \dfrac{x^4 - 3}{6x}$
+> Find the distance from the point ${\frac{1}{2}, \frac{49}{48}}$ to the point $(5, \frac{314}{15})$ along the curve $y = \dfrac{x^4 - 3}{6x}$.
+> *note*: The complete evaluation of this problem is more work than typically required, and is only done for demonstration purposes.
1. $y' = \dfrac{4x^3(6x) - (x^4 + 3)6}{36x^2}$: Find the derivative of the curve using the quotient rule
2. $= \dfrac{18x^4 - 18}{36x^2}$: Simplify
3. $= \dfrac{18(x^4 - 1)}{18(2x^2)}$: Factor out $18$
@@ -169,12 +170,12 @@ $$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$
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. $= \int_{1/2}^5 \sqrt{\dfrac{4x^4 + x^8 -2x^4 + 1}{4x^4}}dx$: Set $1 = \dfrac{4x^4}{4x^4}$ and add
-8. $= \int_{1/2}^5 \sqrt{\dfrac{x^8 + 2x^4 + 1}{4x^4}}dx$: Factor the numerat
-9. $= \int_{1/2}^5 \sqrt{\dfrac{(4x+1)^2}{4x^4}}dx$
-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. $= \frac 1 2 \int_{1/2}^5 (x^2 - x^{-2})dx$
-14. $= \dfrac{1}{2} (\frac{1}{3}x^3 - x^-1)\Big|_{1/2}^5$
+8. $= \int_{1/2}^5 \sqrt{\dfrac{x^8 + 2x^4 + 1}{4x^4}}dx$: Factor the numerator
+9. $= \int_{1/2}^5 \sqrt{\dfrac{(4x+1)^2}{4x^4}}dx$ : Get rid of the square root
+10. = $\int_{1/2}^5 \dfrac{x^4 + 1}{2x^2}dx$: Move the constant $\frac{1}{2}$ outside of the integral
+11. $= \frac{1}{2}\int_{1/2}^5 \dfrac{x^4 + 1}{x^2}$: Rewrite to remove the fraction
+12. $= \frac{1}{2} \int_{1/2}^5 (x^4 + 1)(x^{-2})dx$: distribute
+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