Comparing integrals using the dominance property

Notice that $f(x) - g(x) = x^2 + 1 - 2x = (x-1)^2$, which is a perfect square and therefore satisfies $(x-1)^2 \geq 0$ for every real $x$. In particular, this inequality holds for all $x$ in $[0,3]$, so $f(x) \geq g(x)$ on the entire interval.

By the dominance property of definite integrals, if $f(x) \geq g(x)$ for all $x$ in $[a,b]$, then $\int_a^b f(x)\,dx \geq \int_a^b g(x)\,dx$. Applying this here with $a=0$, $b=3$, we conclude that $\int_0^3 (x^2+1)\,dx \geq \int_0^3 2x\,dx$, so the first integral is larger (they cannot be equal since $f(x) > g(x)$ except at the single point $x=1$).