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  "body": "\n## Proof\n\nWe expand $f(x)$ around a moving point $t$, differentiate in $t$, and integrate back with the fundamental theorem of calculus. The source uses repeated integration by parts; this route packs those steps into one telescoping sum.\n\n1. Let $\\alpha \u003c \\beta$ be the endpoints of $J$, so $\\set{\\alpha, \\beta} = \\set{a, x}$. The hypothesis means that $f, f', \\dots, f^{(k+1)}$ exist on $J$, with [[one-sided derivatives|0HXEX0DPRA]] at $\\alpha$ and $\\beta$.\n2. For $t \\in J$ define $F(t) = \\sum_{j=0}^{k} \\dfrac{f^{(j)}(t)}{j!} (x - t)^j$.\n3. For $j \\le k$, $f^{(j)}$ is differentiable on $J$, so it is continuous on $J$. Inside $J$ this is [[differentiability implies continuity|07ZMX4ANJS]].\n4. At an endpoint, write $f^{(j)}(s) - f^{(j)}(\\alpha) = (s - \\alpha) \\cdot \\dfrac{f^{(j)}(s) - f^{(j)}(\\alpha)}{s - \\alpha}$. The quotient has a finite limit, so the product tends to $0$. The endpoint $\\beta$ is similar.\n5. So $F$ is continuous on $J$, as a sum of products of continuous functions.\n6. By the [[power rule|0JZGW75C5Z]] and the [[chain rule|0RXX2PT1MQ]], $\\dfrac{d}{dt} \\dfrac{(x - t)^j}{j!} = -\\dfrac{(x - t)^{j-1}}{(j-1)!}$ for $j \\ge 1$.\n7. For $0 \\le j \\le k$ and $t \\in J$ let $A_j(t) = \\dfrac{f^{(j+1)}(t)}{j!} (x - t)^j$. Here [[$f^{(j+1)}$|084NGSTWM7]] exists on $J$ by step 1.\n8. By the [[product rule|0V9A43H7WV]] and step 6, the $j$-th term of $F$ has derivative $A_j - A_{j-1}$ on $(\\alpha, \\beta)$ for $j \\ge 1$. The term with $j = 0$ is $f$, with derivative $A_0$.\n9. Adding, the sum telescopes: $F'(t) = A_0 + \\sum_{j=1}^{k} (A_j - A_{j-1}) = A_k(t)$ on $(\\alpha, \\beta)$.\n10. The function $A_k(t) = \\dfrac{f^{(k+1)}(t)}{k!} (x - t)^k$ is continuous on $J$, since $f^{(k+1)}$ is continuous on $J$ by hypothesis.\n11. By the [[fundamental theorem of calculus|0NHGY0N0JB]] with steps 5, 9 and 10, $\\int_\\alpha^\\beta A_k(t) \\, dt = F(\\beta) - F(\\alpha)$.\n12. For $x \u003c a$ the [[integral|0QM4TXT3DR]] $\\int_a^x$ means $-\\int_x^a$. With this convention, step 11 gives $\\int_a^x A_k(t) \\, dt = F(x) - F(a)$ in both cases.\n13. At $t = x$, every term of $F$ with $j \\ge 1$ vanishes, so $F(x) = f(x)$. At $t = a$, $F(a) = P_k(x)$, the value at $x$ of the [[Taylor polynomial|0HFSFVMWQB]] of $f$ at $a$.\n14. So $f(x) - P_k(x) = \\int_a^x \\dfrac{f^{(k+1)}(t)}{k!} (x - t)^k \\, dt$, which is the claim.\n\n## Idea\n\nThe function $F(t)$ is the Taylor polynomial of degree $k$ at $t$, evaluated at $x$. It equals $P_k(x)$ at $t = a$ and $f(x)$ at $t = x$. Its derivative telescopes to the integrand of the remainder, so the remainder is the integral of $F'$ from $a$ to $x$.\n",
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