[proofplan] We construct the canonical antiderivative of $f$ using [FTC Part One](/theorems/190), then compare it with the given antiderivative $G$. Since the two functions have the same [derivative](/page/Derivative) on the interval, their difference has derivative zero. The [Mean Value Theorem](/theorems/186) then forces this difference to be constant, and evaluating at $a$ determines the constant. [/proofplan] [step:Construct the [integral](/page/Integral) antiderivative from FTC Part One] Since $f: [a,b] \to \mathbb{R}$ is [continuous](/page/Continuity) and $a < b$, FTC Part One applies on the interval $[a,b]$. Define the [function](/page/Function) \begin{align*} F: [a,b] &\to \mathbb{R} \\ x &\mapsto \int_a^x f(t)\, d\mathcal{L}^1(t). \end{align*} Then $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and satisfies \begin{align*} F'(x) = f(x) \end{align*} for every $x \in (a,b)$. [/step] [step:Show that the two antiderivatives differ by a constant] Define the function \begin{align*} H: [a,b] &\to \mathbb{R} \\ x &\mapsto G(x) - F(x). \end{align*} Since $G$ and $F$ are continuous on $[a,b]$ and differentiable on $(a,b)$, the difference rule for derivatives gives \begin{align*} H'(x) = G'(x) - F'(x) = f(x) - f(x) = 0 \end{align*} for every $x \in (a,b)$. We prove that $H$ is constant. Let $x \in [a,b]$. If $x = a$, then $H(x) = H(a)$. If $a < x \leq b$, then $H$ is continuous on $[a,x]$ because $G$ and $F$ are continuous on $[a,b]$, and $H$ is differentiable on $(a,x)$ because $G$ and $F$ are differentiable on $(a,b)$. Hence the Mean Value Theorem applied to $H$ on $[a,x]$ gives a point $c \in (a,x)$ such that \begin{align*} H(x) - H(a) = H'(c)(x-a). \end{align*} Since $H'(c) = 0$, we obtain \begin{align*} H(x) - H(a) = 0. \end{align*} Thus $H(x) = H(a)$ for every $x \in [a,b]$, so $H$ is constant on $[a,b]$. [guided] The purpose of introducing $H$ is to isolate the difference between the given antiderivative $G$ and the antiderivative $F$ produced by FTC Part One. Define \begin{align*} H: [a,b] &\to \mathbb{R} \\ x &\mapsto G(x) - F(x). \end{align*} The function $F$ used here is the integral antiderivative \begin{align*} F: [a,b] &\to \mathbb{R} \\ x &\mapsto \int_a^x f(t)\, d\mathcal{L}^1(t). \end{align*} By FTC Part One, $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and satisfies $F'(x) = f(x)$ for every $x \in (a,b)$. The hypothesis on $G$ says that $G$ is continuous on $[a,b]$, differentiable on $(a,b)$, and satisfies $G'(x)=f(x)$ for every $x \in (a,b)$. Hence the difference rule applies on $(a,b)$. For every $x \in (a,b)$, \begin{align*} H'(x) = G'(x) - F'(x) = f(x) - f(x) = 0. \end{align*} Now we use the Mean Value Theorem to convert the derivative information into a statement about function values. Fix $x \in [a,b]$. If $x = a$, then $H(x) = H(a)$ directly. If $a < x \leq b$, we apply the Mean Value Theorem to $H$ on the interval $[a,x]$. Its hypotheses are satisfied: $H$ is continuous on $[a,x]$ because $G$ and $F$ are continuous on $[a,b]$, and $H$ is differentiable on $(a,x)$ because $G$ and $F$ are differentiable on $(a,b)$. Hence there exists $c \in (a,x)$ such that \begin{align*} H(x) - H(a) = H'(c)(x-a). \end{align*} Since $H'(c) = 0$, this becomes \begin{align*} H(x) - H(a) = 0. \end{align*} Therefore $H(x) = H(a)$ for every $x \in [a,b]$. This proves that $H$ is constant on $[a,b]$. [/guided] [/step] [step:Evaluate the constant at $a$ and compute the integral over $[a,b]$] By definition of $F$, \begin{align*} F(a) = \int_a^a f(t)\, d\mathcal{L}^1(t) = 0. \end{align*} Since $H$ is constant on $[a,b]$, \begin{align*} H(b) = H(a). \end{align*} Using the definition $H = G - F$, this gives \begin{align*} G(b) - F(b) = G(a) - F(a) = G(a). \end{align*} Therefore \begin{align*} F(b) = G(b) - G(a). \end{align*} Finally, by the definition of $F$ at $b$, \begin{align*} \int_a^b f(x)\, d\mathcal{L}^1(x) = F(b) = G(b) - G(a). \end{align*} This is the asserted identity. [/step]