From 75151ab5d4c926ab8fdbc10e0c490e87b853a49b Mon Sep 17 00:00:00 2001 From: arc Date: Tue, 1 Apr 2025 10:13:23 -0600 Subject: [PATCH] vault backup: 2025-04-01 10:13:23 --- education/math/MATH1210 (calc 1)/Integrals.md | 6 ++++++ 1 file changed, 6 insertions(+) diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md index 063265c..a071c65 100644 --- a/education/math/MATH1210 (calc 1)/Integrals.md +++ b/education/math/MATH1210 (calc 1)/Integrals.md @@ -98,3 +98,9 @@ $$ \dfrac{d}{dx}\int_0^{\ln{x}}\tan(t) = \tan(\ln(x))*\dfrac{1}{x} $$ $$ F(x) = \int_4^x 2t \space dt = t^2 \Big|_4^x = x^2 - 16$$ > $x$ in top and bottom $$ \dfrac{d}{dx} \int_{2x}^{3x} \sin(t) dt = \dfrac{d}{dx} -\cos(t)\Big|_{2x}^{3x} = \dfrac{d}{dx} (-\cos(3x) + cos(2x) = 3\sin(3x) - 2\sin(2x) $$ + +# The Mean Value Theorem for Integrals +If $f(x)$ is continuous over an interval $[a, b]$ then there is at least one point $c$ in the interval $[a, b]$ such that: +$$f(c) = \dfrac{1}{b-a}\int_a^bf(x)dx $$ +This formula can also be stated as $\int_a^b f(x)dx = f(c)(b-a)$ +This theorem tells us that a continuous function on the closed interval will obtain its average for at least one point in the interval. \ No newline at end of file