 Solution

Suppose the continuous nonnegative function has $f(t_0)>0$ somewhere. Continuity supplies a positive-length interval within $[a,b]$ on which $f\ge f(t_0)/2$; at an endpoint use a one-sided interval. By monotonicity of the Riemann integral, its contribution to the integral is strictly positive, a contradiction. Therefore $\boxed{f\equiv 0}$ under the stated continuous-function assumptions.

For the general integrable case, choose $c\in (a,b)$ and let $f(c)=1$ while $f(t)=0$ for $t\ne c$. This nonnegative indicator function is Riemann integrable with integral zero: isolate $c$ in partition intervals of arbitrarily small total length, making the upper sum arbitrarily small while the lower sum stays zero. Yet $f(c)=1$. This is the single-point changes preserve Riemann integrals phenomenon and supplies the counterexample.