Additivity over adjacent intervals

By the additivity property of definite integrals, for $a < b < c$ we have $\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx$. Here $a=0$, $b=3$, $c=5$, so $\int_0^5 f(x)\,dx = \int_0^3 f(x)\,dx + \int_3^5 f(x)\,dx$.

Substituting the known values gives $8 = \int_0^3 f(x)\,dx + 2$. Solving for the unknown integral yields $\int_0^3 f(x)\,dx = 6$.