Let $$f : \mathbf{R} \to \mathbf{R}$$ be continuous function satisfying $$f(x) + f(x+k) = n$$, for all $$x \in \mathbf{R}$$ where $$k > 0$$ and $$n$$ is a positive integer. If $$I_1 = \int_0^{4nk} f(x) dx$$ and $$I_2 = \int_{-k}^{3k} f(x) dx$$, then

We are given that $$f(x) + f(x+k) = n$$ for all $$x \in \mathbf{R}$$, where $$k > 0$$ and $$n$$ is a positive integer.

Establish the periodicity:

Replacing $$x$$ with $$x+k$$: $$f(x+k) + f(x+2k) = n$$.

Subtracting from the original equation: $$f(x) - f(x+2k) = 0$$, so $$f(x+2k) = f(x)$$.

Therefore $$f$$ is periodic with period $$2k$$.

Key integral property:

Over one period of length $$2k$$:

$$\int_0^{2k} f(x)\,dx = \int_0^k f(x)\,dx + \int_0^k f(x+k)\,dx = \int_0^k [f(x) + f(x+k)]\,dx = \int_0^k n\,dx = nk$$

Compute $$I_1$$:

$$I_1 = \int_0^{4nk} f(x)\,dx$$

Since $$f$$ has period $$2k$$, the interval $$[0, 4nk]$$ contains $$\frac{4nk}{2k} = 2n$$ complete periods.

$$I_1 = 2n \cdot nk = 2n^2k$$

Compute $$I_2$$:

$$I_2 = \int_{-k}^{3k} f(x)\,dx$$

The interval length is $$4k = 2 \cdot 2k$$, which is 2 complete periods.

$$I_2 = 2 \cdot nk = 2nk$$

Check each option:

We have $$I_1 = 2n^2k$$ and $$I_2 = 2nk$$.

Option C: $$I_1 + nI_2 = 2n^2k + n \cdot 2nk = 2n^2k + 2n^2k = 4n^2k$$ $$\checkmark$$

The correct answer is Option C: $$I_1 + nI_2 = 4n^2k$$.