[proofplan] We prove the integral remainder formula by induction on $n$. The base case $n = 0$ is the [Fundamental Theorem of Calculus](/theorems/632). For the inductive step, we assume the formula holds at order $n-1$ and apply [integration by parts](/theorems/210) to the remainder integral, peeling off one more Taylor term and advancing the remainder to order $n$. [/proofplan] [step:Establish the base case $n = 0$ via the Fundamental Theorem of Calculus] For $n = 0$, the formula states \begin{align*} f(a+h) = f(a) + \int_{a}^{a+h} f'(t) \, dt, \end{align*} which is the [Fundamental Theorem of Calculus](/theorems/632) applied to $f'$. This holds because $f'$ is [continuous](/page/Continuity) on $[a, a+h]$ by hypothesis (we assume $n+1 = 1$ continuous [derivatives](/page/Derivative)), and $f$ is an antiderivative of $f'$. [/step] [step:Apply the inductive hypothesis at order $n-1$] Assume the formula holds for $n-1$, i.e., \begin{align*} f(a+h) = \sum_{k=0}^{n-1} \frac{h^k}{k!} f^{(k)}(a) + \int_{a}^{a+h} \frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dt. \end{align*} We must show the remainder integral equals $\frac{h^n}{n!} f^{(n)}(a) + \int_{a}^{a+h} \frac{(a+h-t)^n}{n!} f^{(n+1)}(t) \, dt$. [/step] [step:Integrate by parts to advance the remainder from order $n-1$ to order $n$] Apply [integration by parts](/theorems/210) to the integral $\int_{a}^{a+h} \frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dt$ with the assignments \begin{align*} u = f^{(n)}(t), \qquad dv = \frac{(a+h-t)^{n-1}}{(n-1)!} \, dt. \end{align*} Then \begin{align*} du = f^{(n+1)}(t) \, dt, \qquad v = -\frac{(a+h-t)^n}{n!}, \end{align*} where $v$ is obtained by integrating: $\int \frac{(a+h-t)^{n-1}}{(n-1)!} \, dt = -\frac{(a+h-t)^n}{n \cdot (n-1)!} = -\frac{(a+h-t)^n}{n!}$. [Integration by parts](/theorems/2098) gives \begin{align*} \int_{a}^{a+h} \frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dt &= \left[-\frac{(a+h-t)^n}{n!} f^{(n)}(t)\right]_{t=a}^{t=a+h} + \int_{a}^{a+h} \frac{(a+h-t)^n}{n!} f^{(n+1)}(t) \, dt. \end{align*} Evaluating the boundary term: \begin{align*} \left[-\frac{(a+h-t)^n}{n!} f^{(n)}(t)\right]_{t=a}^{t=a+h} &= -\frac{0^n}{n!} f^{(n)}(a+h) + \frac{h^n}{n!} f^{(n)}(a) = \frac{h^n}{n!} f^{(n)}(a). \end{align*} Therefore \begin{align*} \int_{a}^{a+h} \frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dt = \frac{h^n}{n!} f^{(n)}(a) + \int_{a}^{a+h} \frac{(a+h-t)^n}{n!} f^{(n+1)}(t) \, dt. \end{align*} [guided] Why do we choose these particular assignments for [integration by parts](/theorems/210)? We set \begin{align*} u = f^{(n)}(t), \qquad dv = \frac{(a+h-t)^{n-1}}{(n-1)!} \, dt, \end{align*} so that $du = f^{(n+1)}(t)\,dt$ (raising the derivative order) and $v = -\frac{(a+h-t)^n}{n!}$ (raising the power of the kernel by one, with a sign flip from the chain rule). The antiderivative is computed by \begin{align*} \int \frac{(a+h-t)^{n-1}}{(n-1)!}\,dt = -\frac{(a+h-t)^n}{n \cdot (n-1)!} = -\frac{(a+h-t)^n}{n!}. \end{align*} The integration by parts formula $\int u\,dv = [uv] - \int v\,du$ then yields \begin{align*} \int_{a}^{a+h} \frac{(a+h-t)^{n-1}}{(n-1)!} f^{(n)}(t)\,dt = \left[-\frac{(a+h-t)^n}{n!} f^{(n)}(t)\right]_{t=a}^{t=a+h} + \int_{a}^{a+h} \frac{(a+h-t)^n}{n!} f^{(n+1)}(t)\,dt. \end{align*} The boundary term at $t = a+h$ vanishes because $(a+h-(a+h))^n = 0^n = 0$. The boundary term at $t = a$ contributes $\frac{h^n}{n!} f^{(n)}(a)$, since $(a+h-a)^n = h^n$. This is exactly the next Taylor coefficient, which is the mechanism by which integration by parts "peels off" one more term of the Taylor polynomial at each inductive step. [/guided] [/step] [step:Substitute back to obtain the formula at order $n$] Substituting the result of the integration by parts into the inductive hypothesis: \begin{align*} f(a+h) &= \sum_{k=0}^{n-1} \frac{h^k}{k!} f^{(k)}(a) + \frac{h^n}{n!} f^{(n)}(a) + \int_{a}^{a+h} \frac{(a+h-t)^n}{n!} f^{(n+1)}(t) \, dt \\ &= \sum_{k=0}^{n} \frac{h^k}{k!} f^{(k)}(a) + \int_{a}^{a+h} \frac{(a+h-t)^n}{n!} f^{(n+1)}(t) \, dt. \end{align*} This completes the induction. [/step]