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  "body": "\n## Proof\n\nWe find an antiderivative and apply the fundamental theorem of calculus.\n\n1. Let $n \\ge 0$ be an integer, and let $F(x) = \\frac{x^{n+1}}{n+1}$ and $f(x) = x^n$ for real $x$, with $x^0 = 1$.\n2. Since $n + 1 \\ge 1$, the [[power rule|0JZGW75C5Z]] says that $x \\mapsto x^{n+1}$ is differentiable on $\\R$ with derivative $(n+1) x^n$.\n3. By the [[constant multiple rule|0Z9JB71BS0]] with the constant $\\frac{1}{n+1}$, $F$ is differentiable on $\\R$ and $F'(x) = x^n = f(x)$.\n4. [[Differentiable functions are continuous|07ZMX4ANJS]], so $F$ is continuous on $\\R$.\n5. The function $f$ is continuous on $\\R$. For $n = 0$ it is the constant $1$. For $n \\ge 1$ it is differentiable by the power rule, so it is continuous by the same result as in step 4.\n6. Assume $a \u003c b$. By step 5, $f$ is [[Riemann integrable on $[a, b]$|06SESV7GQY]].\n7. By steps 3, 4 and 6, $F$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $F' = f$ there with $f$ integrable. The [[second fundamental theorem of calculus|0XYAFCN0PT]] gives $\\int_a^b x^n \\, dx = F(b) - F(a)$.\n8. Assume $a = b$. By the convention for the [[definite integral|021WEM680W]], $\\int_a^a x^n \\, dx = 0 = F(a) - F(a)$.\n9. Assume $a \u003e b$. By the same convention and step 7 on $[b, a]$, $\\int_a^b x^n \\, dx = -\\int_b^a x^n \\, dx = -(F(a) - F(b)) = F(b) - F(a)$.\n10. In all cases, $\\int_a^b x^n \\, dx = F(b) - F(a) = \\frac{b^{n+1} - a^{n+1}}{n+1}$.\n\n## Idea\n\nThe power rule shows that $x^{n+1}/(n+1)$ has derivative $x^n$. The fundamental theorem of calculus then turns the integral into a difference of values of this antiderivative. The sign convention for reversed limits makes the formula hold for every order of $a$ and $b$.\n",
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