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  "body": "\n## Proof\n\nWe apply the second fundamental theorem to the product $u v$.\n\n1. Let $a \u003c b$ and let $u, v$ be [[continuously differentiable|06SKCHDQT5]] on $[a, b]$. On a closed interval we read this in the usual way.\n   So $u$ and $v$ are differentiable on $(a, b)$, have [[one-sided derivatives|0HXEX0DPRA]] at $a$ and $b$, and $u'$ and $v'$ are continuous on $[a, b]$.\n2. The functions $u$ and $v$ are continuous on $[a, b]$. Inside, this holds because [[differentiable functions are continuous|07ZMX4ANJS]].\n   At $a$, write $u(a + h) - u(a) = h \\cdot \\frac{u(a + h) - u(a)}{h}$. As $h \\to 0^+$ the quotient tends to $u'(a)$ and $h$ tends to $0$, so $u(a + h) \\to u(a)$. The point $b$ and the function $v$ are the same.\n3. Let $w = u v$. Then $w$ is continuous on $[a, b]$, as a [[product of continuous functions|0VGMHCF4SC]].\n4. By the [[product rule|0V9A43H7WV]], $w$ is differentiable on $(a, b)$ with $w' = u' v + u v'$.\n5. The functions $u' v$ and $u v'$ are products of continuous functions on $[a, b]$, so they are continuous. Since [[continuous functions are Riemann integrable|06SESV7GQY]], both are integrable on $[a, b]$.\n6. By the [[linearity of the integral|080RWVVSB6]], $g = u' v + u v'$ is integrable on $[a, b]$, and\n   $$\\int_a^b g(x) \\, dx = \\int_a^b u'(x) v(x) \\, dx + \\int_a^b u(x) v'(x) \\, dx.$$\n7. The function $g$ is integrable on $[a, b]$ and equals $w'$ on $(a, b)$. By the [[second fundamental theorem of calculus|0XYAFCN0PT]], $\\int_a^b g(x) \\, dx = w(b) - w(a) = u(b) v(b) - u(a) v(a)$.\n8. Combining steps 6 and 7 and moving the integral of $u' v$ to the other side gives\n   $$\\int_a^b u(x) v'(x) \\, dx = u(b) v(b) - u(a) v(a) - \\int_a^b u'(x) v(x) \\, dx.$$\n\n## Idea\n\nThe product rule says that $u v$ is an antiderivative of $u' v + u v'$. Integrating this sum with the fundamental theorem gives the boundary term $u v$ at $b$ minus at $a$. Splitting the integral of the sum moves the derivative from one factor to the other.\n",
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