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.
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