Let $f(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}$, where $\mathrm{g}$ is a non-zero
even function. If $f(\mathrm{x}+5)=\mathrm{g}(\mathrm{x})$, then $\int_{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$ equals-
Correct Option: 1
$f(x)=\int_{0}^{x} g(t) d t$
$f(-x)=\int_{0}^{-x} g(t) d t$
put $\mathrm{t}=-\mathrm{u}$
$=-\int_{0}^{\mathrm{x}} \mathrm{g}(-\mathrm{u}) \mathrm{du}$
$=-\int_{0}^{x} g(u) d(u)=-f(x)$
$\Rightarrow f(-\mathrm{x})=-f(\mathrm{x})$
$\Rightarrow f(\mathrm{x})$ is an odd function
Also $f(5+x)=g(x)$
$f(5-x)=g(-x)=g(x)=f(5+x)$
$\Rightarrow f(5-\mathrm{x})=f(5+\mathrm{x})$
Now
$\mathrm{I}=\int_{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$
$\mathrm{t}=\mathrm{u}+5$
$I=\int_{-5}^{x-5} f(u+5) d u$
$=\int_{-5}^{x-5} g(u) d u$
$=\int_{-5}^{x-5} f^{\prime}(u) d u$
$=f(\mathrm{x}-5)-f(-5)$
$=-f(5-\mathrm{x})+f(5)$
$=f(5)-f(5+\mathrm{x})$
$=\int_{5+x}^{5} f^{\prime}(t) d t=\int_{5+x}^{5} g(t) d t$