 Solution

Use the ring homomorphism $k[t]\to k[x]$ given by $t\mapsto x^2$. The source scheme $X=\mathrm{Spec}k[x]$ is an integral scheme. At the rational point $y=(t)$, its residue field is $k$, and its scheme-theoretic fibre isThe class of $x$ is nonzero but has square zero, so this is not a reduced scheme. This example works over every field. An integral total space therefore need not have reduced fibres.