= Solution
{c}
For $a<b$, choose the uniform <partition of an interval> into $n$ pieces and its midpoint tags
$$
\xi_i=a+\left(i-\frac12\right)\frac{b-a}{n},\qquad1\leq i\leq n.
$$
Their <arithmetic mean> is exactly $(a+b)/2$. Apply the given <Jensen inequality> with equal weights $\lambda_i=1/n$:
$$
\frac1n\sum_{i=1}^n f(\xi_i)\leq f\left(\frac1n\sum_{i=1}^n\xi_i\right)=f\left(\frac{a+b}2\right).
$$
Multiplying by $b-a$ gives a midpoint <Riemann sum> on the left. Its <mesh of a partition> is $(b-a)/n\to0$, so the <Riemann sum convergence theorem> gives
$$
\boxed{\int_a^b f(x)\,dx\leq(b-a)f\left(\frac{a+b}2\right).}
$$
For $a=b$ the equality is immediate. Together with part <7/c/i>, this is the <Hermite-Hadamard inequality> with the directions appropriate to a <concave function>.
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