Interesting Integrals

Let us write the above integral as the following:

$\int_1^3 f(x) dx - \int_1^3 g(x) dx = [\int_{-1}^3 f(x) dx - \int_{-1}^0 f(x) dx - \int_0^1 f(x) dx] - [\int_{-1}^3 g(x) dx - \int_{-1}^0 g(x) dx - \int_0^1 g(x) dx]$ (i).

Since $\int_{-1}^0 f(x) dx = \int_{-1}^0 g(x) dx$ and $\int_{-1}^3 f(x) dx = \int_{-1}^3 g(x) dx$ , we can now reduce (i) to:

$\int_1^3 f(x) dx - \int_1^3 g(x) dx = \int_0^1 g(x) dx - \int_0^1 f(x) dx$ (ii)

Since $g(x)$ is even and $f(x)$ is odd, we now have:

$\int_0^1 g(x) dx = \int_{-1}^0 g(x) dx = 2$ and $-\int_0^1 f(x) dx = \int_{-1}^0 f(x) dx = 2$

After substituting these values into (ii), we finally arrive at the result:

$\int_1^3 f(x) dx - \int_1^3 g(x) dx = \int_0^1 g(x) dx - \int_0^1 f(x) dx = 2 + 2 = \boxed{4}.$