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education/math/MATH1210 (calc 1)/Integrals.md
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@ -87,8 +87,10 @@ Average = $\dfrac{1}{b-a} \int_a^b f(x)dx$
Let $f$ be a continuous function on the closed interval $[a, b]$ and let $F$ be any antiderivative of $f$, then:
$$\int_a^b f(x) dx = F(b) - F(a)$$
## Proof
## Mean Value Theorem
Between the interval $[a, b]$, the average rate of change of a function **must be equal** to at least one point between $[a, b]$:
$$ f'(c) = \dfrac{f(b) - f(a)}{x-a} $$
For some $c$ between $a$ and $b$.
For some $c$ between $a$ and $b$.
1. $F(b) - F(a) = F(x_1) - F(x_0)$
2. $= (F(x_1) - F(x_0) + (F(x_2) - F(x_1) + (F(X_3) - F(X_2) + \cdots + (F(x_n) - f(x_{n - 1})$ 3. $=