= Solution
It suffices to assume $f\in C^2[a,b]$. The <polynomial interpolation error> for the endpoint <linear interpolant> gives, for every interior $x$, some $\xi_x\in(a,b)$ with
$$
f(x)-\ell(x)=\frac{f''(\xi_x)}2(x-a)(x-b).
$$
Let $M=\max_{[a,b]}|f''|$, which exists by <continuity> of the <second derivative> and the <extreme value theorem>. Since $(x-a)(b-x)\geq0$,
$$
|f(x)-\ell(x)|\leq\frac M2(x-a)(b-x).
$$
Integrating this pointwise inequality and using the <triangle inequality for an integral> yields
$$
\begin{aligned}
\left|\int_a^bf(x)\,dx-T(f)\right|
&\leq\frac M2\int_a^b(x-a)(b-x)\,dx\\
&=\frac M2\int_0^{b-a}t\bigl((b-a)-t\bigr)\,dt
=\frac{M(b-a)^3}{12}.
\end{aligned}
$$
Thus
$$
\boxed{|I(f)-T(f)|\leq\frac{(b-a)^3}{12}\max_{[a,b]}|f''|.}
$$
This derivation does not require choosing $\xi_x$ as a continuous <function> of $x$; the pointwise upper bound is sufficient.
The constant is sharp. Take the <quadratic polynomial> $f(x)=(x-a)^2$ and write $L=b-a$. Its exact <integral> is $L^3/3$, its <trapezoidal rule> value is $L^3/2$, and $\max|f''|=2$. Consequently
$$
|I(f)-T(f)|=\frac{L^3}{6}=\frac{L^3}{12}\max|f''|.
$$
Equality holds, so \b[no smaller universal constant than $1/12$ works] under these assumptions. This is the <sharp one-panel trapezoidal error bound>.
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