= Solution
The <heat equation> and its homogeneous <Dirichlet boundary conditions> are <linear>, so <superposition> of the separated modes gives
$$
\boxed{u(x,t)=\sum_{n=1}^{\infty}\alpha_n\sin\frac{n\pi x}{L}
\exp\left(-\frac{Dn^2\pi^2t}{L^2}\right).}
$$
The <Fourier sine basis> represents the initial profile on the interior of the interval. At $t=0$ its <Fourier sine series> must equal $x(1-x)$. Multiplying by $\sin(m\pi x/L)$, integrating from $0$ to $L$, and using <orthogonality> gives
$$
\int_0^L\sin\frac{n\pi x}{L}\sin\frac{m\pi x}{L}\,dx
=\begin{cases}L/2,&n=m,\\0,&n\ne m.\end{cases}
$$
Therefore the required <Fourier sine coefficients> are
$$
\boxed{\alpha_n=\frac2L\int_0^L x(1-x)\sin\frac{n\pi x}{L}\,dx,
\qquad n=1,2,\ldots.}
$$
For every $t>0$, the exponential factors ensure convergence of the differentiated series, and every term satisfies the <partial differential equation> and the <boundary conditions>. The <Fourier sine series> recovers the smooth initial profile at every $0<x<L$ as $t\downarrow0$.
The scan explicitly prints $x(1-x)$ on $(0,L)$; it must not silently become $x(L-x)$. For $L\ne1$, the interior profile tends to $L(1-L)\ne0$ at the right endpoint, whereas the prescribed <boundary condition> is zero for $t>0$. The solution is therefore not continuous at the corner $(L,0)$, although it satisfies every condition on the printed open ranges. A solution continuous on the closed space-time domain with these data would require $L=1$.<Discrete mathematics, continuous mathematics, and probability examination (2015) by University of Oxford>[Printed page 7, Question 6(b), initial profile and ranges]{ref}
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