arc/notes
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

vault backup: 2025-04-01 10:13:23

Browse Source
This commit is contained in:
arc 2025-04-01 10:13:23 -06:00
parent 5463d89d1c
commit 75151ab5d4
1 changed files with 6 additions and 0 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
education/math/MATH1210 (calc 1)/Integrals.md
View File

@ -98,3 +98,9 @@ $$ \dfrac{d}{dx}\int_0^{\ln{x}}\tan(t) = \tan(\ln(x))*\dfrac{1}{x} $$
$$ F(x) = \int_4^x 2t \space dt = t^2 \Big|_4^x = x^2 - 16$$ $$ F(x) = \int_4^x 2t \space dt = t^2 \Big|_4^x = x^2 - 16$$
> $x$ in top and bottom > $x$ in top and bottom
$$ \dfrac{d}{dx} \int_{2x}^{3x} \sin(t) dt = \dfrac{d}{dx} -\cos(t)\Big|_{2x}^{3x} = \dfrac{d}{dx} (-\cos(3x) + cos(2x) = 3\sin(3x) - 2\sin(2x) $$ $$ \dfrac{d}{dx} \int_{2x}^{3x} \sin(t) dt = \dfrac{d}{dx} -\cos(t)\Big|_{2x}^{3x} = \dfrac{d}{dx} (-\cos(3x) + cos(2x) = 3\sin(3x) - 2\sin(2x) $$
# 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.