I'm undergoing an online course and this question is asked:
Suppose that the definite integral $\int_3^{3.09}3.09 f(x)\,dx =0.81$.
Approximate $f(3)$ as well as you can given this information.
The only examples given include more information, such as f(n)=a.
The idea is that if $\int_3^{3.09}f(x)dx=0.81$, then an antiderivative $F(x)$ of $f(x)$ satisfies $$ F(3.09)-F(3)=0.81. $$ The Mean Value Theorem tells you that there exists some number $c$ between $3$ and $3.09$ such that $$ \frac{F(3.09)-F(3)}{3.09-3}=F'(c)=f(c). $$ Assuming that $f$ is continuous, $f(c)$ cannot be too different from $f(3)$. So we have the approximation $$ f(3)\simeq\frac{F(3.09)-F(3)}{3.09-3}=\frac{0.81}{0.09}=9. $$