Finding the integral without knowing the actual equation, but some several information is given. For any real number x, $f(x) = f(x + 2)$.
Let $f(x) = -x^2 + 2, -1 \le x \le  1$
I need to compute: $$\int_{-1}^6 f(x) \,dx$$
I came up with the idea that I might split the range $(-1, 6)$ to its components. Like:
$$\int_{-1}^6 f(x) \,dx = \int_{-1}^1 f(x) \,dx + \int_{1}^6 f(x) \,dx$$
By this action, we can figure out the 1st term, but I have no idea what to do with the 2nd term. I am stuck, could you please help me to complete with this idea, or if you have your own approach that is welcomed too.

For any real number $\ x,\ f(x) = f(x + 2).\ $ Then use for example,
$$\int_{-1}^1 f(x) \,dx = \int_{1}^3 f(x) \,dx = \int_{3}^5 f(x) \,dx,$$
and
$$\int_{-1}^6 f(x) \,dx = \int_{-1}^1 f(x) \,dx + \int_{1}^3 f(x) \,dx + \int_{3}^5 f(x) \,dx + \int_{5}^6 f(x) \,dx$$
