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Let $f(x)$ be an even function with period 2 and $f(x)$ be integrable on every interval. If $g(x)=\int_0^x f(t) d t$, then $g(x+2)=$
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Verified Answer
The correct answer is:
$g(x)+g(2)$
We have,
$$
\begin{aligned}
& \qquad g(x)=\int_0^x f(t) d t \\
& \quad g(x+2)=\int_0^{x+2} f(t) d t \\
& \int_0^2 f(t) d t+\int_2^{2+x} f(t) d t \\
& \quad g(x+2)=\int_0^2 f(t) d t+\int_2^x f(t) d t[\because f(x) \text { is periodic } \\
& \text { with period }=2] \\
& g(x+2)=g(2)+g(x)(\because f(x), \text { is even function } \\
& \text { periodic with } 2) \\
& \therefore \quad g(x+2)=g(x)+g(2)
\end{aligned}
$$
$$
\begin{aligned}
& \qquad g(x)=\int_0^x f(t) d t \\
& \quad g(x+2)=\int_0^{x+2} f(t) d t \\
& \int_0^2 f(t) d t+\int_2^{2+x} f(t) d t \\
& \quad g(x+2)=\int_0^2 f(t) d t+\int_2^x f(t) d t[\because f(x) \text { is periodic } \\
& \text { with period }=2] \\
& g(x+2)=g(2)+g(x)(\because f(x), \text { is even function } \\
& \text { periodic with } 2) \\
& \therefore \quad g(x+2)=g(x)+g(2)
\end{aligned}
$$
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