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  "body": "\n## Statement\n\nLet $a \u003c b$ be real numbers and let $f \\colon [a, b] \\to \\R$ be a [[convex function|0VV0HSW1ZS]].\nThen $f$ is Riemann integrable on $[a, b]$ and\n$$f\\Big( \\frac{a + b}{2} \\Big) \\le \\frac{1}{b - a} \\int_a^b f(x) \\, dx \\le \\frac{f(a) + f(b)}{2}.$$\n\n## Remarks\n\nProof idea: a convex function on $[a, b]$ is bounded and is continuous on $(a, b)$, so it is integrable.\nFor the left inequality, convexity gives $f\\big(\\frac{a+b}{2}\\big) \\le \\frac{1}{2}\\big(f(x) + f(a + b - x)\\big)$; integrate over $x \\in [a, b]$.\nFor the right inequality, the graph lies below the chord: $f(x) \\le \\frac{b - x}{b - a} f(a) + \\frac{x - a}{b - a} f(b)$; integrate this bound, whose mean value is $\\frac{f(a) + f(b)}{2}$.\n\nExample: $f(x) = x^2$ on $[0, 1]$ gives $\\frac{1}{4} \\le \\frac{1}{3} \\le \\frac{1}{2}$.\n\nFor a concave function both inequalities are reversed.\n\n## Explanations\n\n### L1 human\n\nThe average value of a convex function over an interval lies between two simple numbers.\nIt is at least the value at the midpoint, and at most the average of the two end values.\nGeometrically, the graph lies above a supporting line at the midpoint and below the chord joining the ends.\n\n### L2 human\n\nThe inequality gives error bounds for the midpoint rule and the trapezoidal rule: for a convex integrand the midpoint rule underestimates the integral and the trapezoidal rule overestimates it.\nIt is a two-sided refinement of [[Jensen's inequality|0RWMRTPC87]] for the uniform distribution on $[a, b]$.\n",
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