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  "body": "\n## Proof\n\nWe find an antiderivative $F$ of $f$ and apply the second fundamental theorem to $F \\circ g$.\n\n1. Let $a \u003c b$. We read [[continuously differentiable|06SKCHDQT5]] on the closed interval $[a, b]$ in the usual way.\n   So $g$ is differentiable on $(a, b)$, has [[one-sided derivatives|0HXEX0DPRA]] at $a$ and $b$, and $g'$ is continuous on $[a, b]$.\n2. The function $g$ is continuous on $[a, b]$. Inside, [[differentiable functions are continuous|07ZMX4ANJS]]. At $a$, $g(a + h) - g(a) = h \\cdot \\frac{g(a + h) - g(a)}{h}$ tends to $0 \\cdot g'(a) = 0$ as $h \\to 0^+$, and $b$ is the same.\n3. By the [[extreme value theorem|0WS1A8E2P9]], $g$ has a minimum $m$ and a maximum $M$ on $[a, b]$. Let $J$ be the interval on which $f$ is continuous. It contains $m$ and $M$, so it contains $[m, M] \\supseteq g([a, b])$.\n4. Let $k(u) = \\min(\\max(u, m), M)$ for $u \\in \\mathbb{R}$. Then $k(u) \\in [m, M]$, and $k(u) = u$ for $u \\in [m, M]$.\n5. We have $\\max(u, m) = \\frac{u + m + |u - m|}{2}$. With the [[reverse triangle inequality|0C9S13WKMF]], $|\\max(u, m) - \\max(v, m)| \\le |u - v|$. In the same way $|\\min(u, M) - \\min(v, M)| \\le |u - v|$.\n   So $|k(u) - k(v)| \\le |u - v|$, and $k$ is [[Lipschitz|0QNVQ4VD9T]]. Hence [[$k$ is continuous|0K1EWAR6AF]].\n6. Let $\\tilde f = f \\circ k$ on $\\mathbb{R}$. It is continuous, as a [[composition of continuous functions|0NRQPQX3VA]], and $\\tilde f = f$ on $[m, M]$.\n7. Let $F(y) = \\int_0^y \\tilde f(s) \\, ds$ for $y \\in \\mathbb{R}$. By the theorem that [[the integral of a continuous function is an antiderivative|0NC2A3HWGT]], $F$ is differentiable on $\\mathbb{R}$ with $F' = \\tilde f$.\n8. Let $H = F \\circ g$ on $[a, b]$. It is continuous by steps 2 and 7, since differentiable functions are continuous and compositions of continuous functions are continuous.\n9. For $x \\in (a, b)$, the [[chain rule|0RXX2PT1MQ]] gives $H'(x) = F'(g(x)) g'(x) = \\tilde f(g(x)) g'(x) = f(g(x)) g'(x)$, since $g(x) \\in [m, M]$.\n10. The function $x \\mapsto f(g(x)) g'(x) = \\tilde f(g(x)) g'(x)$ is continuous on $[a, b]$, by [[products of continuous functions|0VGMHCF4SC]]. So it is [[Riemann integrable|06SESV7GQY]], and it extends $H'$.\n11. By the [[second fundamental theorem of calculus|0XYAFCN0PT]], $\\int_a^b f(g(x)) g'(x) \\, dx = H(b) - H(a) = F(g(b)) - F(g(a))$.\n12. Let $\\alpha = g(a)$ and $\\beta = g(b)$. Both lie in $[m, M]$, so $f = \\tilde f$ on the closed interval between them. Their [[Riemann sums|0HBY48G8G1]] there agree, so their integrals agree.\n13. If $\\alpha \u003c \\beta$, apply the second fundamental theorem to $F$ on $[\\alpha, \\beta]$. $F$ is continuous and differentiable with $F' = \\tilde f$, and $\\tilde f$ is continuous, hence integrable. So $\\int_\\alpha^\\beta f(u) \\, du = F(\\beta) - F(\\alpha)$.\n14. If $\\alpha \u003e \\beta$, the same argument on $[\\beta, \\alpha]$ and the sign convention for the [[definite integral|021WEM680W]] give $\\int_\\alpha^\\beta f(u) \\, du = -(F(\\alpha) - F(\\beta)) = F(\\beta) - F(\\alpha)$.\n    If $\\alpha = \\beta$, both sides are $0$.\n15. By steps 11 to 14, $\\int_a^b f(g(x)) \\, g'(x) \\, dx = F(g(b)) - F(g(a)) = \\int_{g(a)}^{g(b)} f(u) \\, du$.\n\n## Idea\n\nIf $F$ is an antiderivative of $f$, the chain rule makes $F \\circ g$ an antiderivative of $(f \\circ g) \\, g'$. So both integrals equal $F(g(b)) - F(g(a))$. The extension $\\tilde f$ only makes sure that $F$ is defined and differentiable on an open set around all values of $g$.\n",
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