From f8803724a6f40a98e2f1bfab995e9142145b525f Mon Sep 17 00:00:00 2001 From: arc Date: Thu, 17 Apr 2025 09:30:41 -0600 Subject: [PATCH] vault backup: 2025-04-17 09:30:41 --- education/math/MATH1210 (calc 1)/Integrals.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md index 89082f7..1640c29 100644 --- a/education/math/MATH1210 (calc 1)/Integrals.md +++ b/education/math/MATH1210 (calc 1)/Integrals.md @@ -109,7 +109,7 @@ This theorem tells us that a continuous function on the closed interval will obt # U-Substitution When you see $dx$ or $du$ in a function, it can be thought of as roughly analogous to $\Delta x$, where the change in $x$ is infinitesimally small. -Thinking back to derivatives, when solving for $\dfrac{dy}{dx}$, you're solving for the rate of change of $y$ (across an infinitely small distance) over the rate of change of $x$ (across an infinitely small i$nstance). Given that the *slope* of a line is described as $\dfrac{\text{rise}}{\text{run}}$, +Thinking back to derivatives, when solving for $\dfrac{dy}{dx}$, you're solving for the rate of change of $y$ (across an infinitely small distance) over the rate of change of $x$ (across an infinitely small instance). Given that the *slope* of a line is described as $\dfrac{\text{rise}}{\text{run}}$, we know that $\dfrac{dy}{dx}$ describes the slope of a line at a particular point. ## Formulas - $\int k {du} = ku + C$ - $\int u^n du = \frac{1}{n+1}u^{n+1} + C$