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(B) Given that $f(x)$ is an even function with period $2$. Since $f(x)$ is periodic with period $2$, we have $\int_0^2 f(t) dt = \int_a^{a+2} f(t) dt$

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$g(x) + g(2)$

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(B) Given that $f(x)$ is an even function with period $2$. Since $f(x)$ is periodic with period $2$, we have $\int_0^2 f(t) dt = \int_a^{a+2} f(t) dt$ for any $a$. Also, for an even function, $\int_{-a}^a f(t) dt = 2 \int_0^a f(t) dt$. Specifically, $\int_{-1}^1 f(t) dt = 2 \int_0^1 f(t) dt$. Since the period is $2$, $\int_0^2 f(t) dt = \int_{-1}^1 f(t) dt = 2 \int_0^1 f(t) dt$. Now, $g(x+2) = \int_0^{x+2} f(t) dt = \int_0^2 f(t) dt + \int_2^{x+2} f(t) dt$. Using the periodicity property $\int_2^{x+2} f(t) dt = \int_0^x f(t) dt = g(x)$. Thus, $g(x+2) = g(2) + g(x)$.

Let $f(x)$ be an even function with period $2$ and $f(x)$ be integrable on every interval. If $g(x) = \int_0^x f(t) dt$, then $g(x+2) =$

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