Integrals with the Delta Function: Do I have to integrate and put the integration limits into the function? Or $\int_1^5 f(x)\delta(x-3) \, dx=f(3)$?

Question

I tried exercise a) from this page and got a different result https://en.universaldenker.org/exercises/1359

$f(x)=2x^2-x+1$

$$\int_1^5 (2x^2-x+1)\delta(x-3) \, dx = \int_{1}^5 f(3) \, dx = \int_{1}^5 2\cdot 3^2-3+1\, dx= \int_{1}^5 16\, dx= [16x]_{1}^5=16\cdot 5-16=64$$

The solution on the website is $f(3)=16$, I don't understand why?