Let . Show that . Find the value of .

Given $\int_1^5 3f(x) dx = 12 \Rightarrow 3\int_1^5 f(x) dx = 12 \Rightarrow \int_1^5 f(x) dx = 4$. Using the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$: $\int_5^1 f(x) dx = -\int_1^5 f(x) dx = -4$. Hence proved.

Using the additive property of definite integrals, $\int_1^2 g(x) dx + \int_2^5 g(x) dx = \int_1^5 g(x) dx$: $\int_1^2 (x + f(x)) dx + \int_2^5 (x + f(x)) dx = \int_1^5 (x + f(x)) dx$ $= \int_1^5 x dx + \int_1^5 f(x) dx$ $= \left[\frac{x^2}{2}\right]_1^5 + 4$ $= \left(\frac{25}{2} - \frac{1}{2}\right) + 4 = 12 + 4 = 16$.