> [!note] Theorem > > Suppose that a [[Science/Math/Functions/Function|function]] is [[Continuity|continuous]] over $[a, b]$: > 1. If $g(x)$ is defined by the [[Definite Integral|definite integral]] $g(x) = \int_{a}^{x}f(t)dt$, then its [[Derivative|derivative]] $g^\prime(x) = f(x)$ is continuous over $[a, b]$ and [[Derivative|differentiable]] over $(a, b)$. > 2. $\int_{a}^{b}f(x)dx = F(b) - F(a)$, where $F$ is any [[Antiderivative|antiderivative]] of $f$. > [!info] Proof > > ### Part 1 > > If $x$ and $x + h$ are in $(a, b)$, then: > $ > \begin{align*} > g(x + h) - g(x) &= \int_{a}^{x + h}f(t)dt - \int_{a}^{x}f(t)dt \\ > &= \int_{a}^{x}f(t)dt + \int_{x}^{x + h}f(t)dt - \int_{a}^{x}f(t)dt \\ > &= \int_{x}^{x + h}f(t)dt > \end{align*} > $ > > And so for $h \ne 0$: > $ > \frac{g(x + h) - g(x)}{h} = \frac{1}{h}\int_{x}^{x + h}f(t)dt > $ > > Since $f$ is continuous over $[x, x + h] \in [a, b]$, by [[Extreme Value Theorem|extreme value theorem]], there exists numbers $u$ and $v$ in $[x, x + h]$, such that $f(u) = m$ and $f(v) = M$, where $m$ and $M$ are the [[Extrema#Absolute|absolute minimum and maximum]] of $f$ over $[x, x + h]$, which means that > > $ > \begin{align*} > mh \le &\int_{x}^{x + h}f(t)dt \le Mh \\ > f(u)h \le &\int_{x}^{x + h}f(t)dt \le f(v)h \\ > f(u) \le &\frac{1}{h}\int_{x}^{x + h}f(t)dt \le f(v) \\ > f(u) \le &\frac{g(x + h) - g(x)}{h} \le f(v) > \end{align*} > $ > > Let $h \to 0$. Then $u \to x$ and $v \to x$, since $u$ and $v$ lie between $x$ and $x + h$, establish the [[Limit|limits]] > $ > \lim_{h \to 0}f(u) = \lim_{u \to x}f(u) = f(x) \quad > \text{and} \quad > \lim_{h \to 0}f(v) = \lim_{v \to x}f(v) = f(x) > $ > Since $f$ is continuous at $x$, $f(u) \le f(x) \le f(v)$, by [[Squeeze Theorem|squeeze theorem]]: > $ > g^\prime(x) = \lim_{h \to 0}\frac{g(x + h) - g(x)}{h} = f(x) > $ > > > ### Part 2 > > Let $g(x) = \int_{a}^{x}f(t)dt$. It's known from part 1 that $g^\prime(x) = f(x)$, which means that $g$ is an antiderivative of $f$. If $F$ is any other antiderivative of $f$ over $[a, b]$, then $F$ and $g$ differ by a constant: > $ > F(x) = g(x) + C > $ > > If $a$ is plugged in for $x$, $x = a$ in the formula for $g(x)$: > $ > g(a) = \int_{a}^{a}f(t)dt = 0 > $ > > If $a$ and $b$ are plugged in for $x$: > $ > \begin{align*} > F(b) - F(a) &= [g(b) + C] - [g(a) + C] \\ > &= g(b) - g(a) = g(b) - 0 \\ > &= g(b) = \int_{a}^{b}f(t)dt > \end{align*} > $