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 $\int^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$
equals-
Correct Option: 1
$f(x)=\int_{0}^{x} g(t) d t$
$f(-\mathrm{x})=\int_{0}^{-\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}$
put $t=-\mathbf{u}$
$=-\int_{0}^{x} g(-u) d u$
$=-\int_{0}^{x} g(u) d(u)=-f(x)$
$\Rightarrow f(-\mathrm{x})=-f(\mathrm{x})$
$\Rightarrow f(\mathrm{x})$ is an odd function
Also ƒ(5 + x) = g(x)
ƒ(5 – x) = g(–x) = g(x) = ƒ(5 + x)
$\Rightarrow f(5-x)=f(5+x)$
Now
$\mathrm{I}=\int_{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$
$\mathrm{t}=\mathrm{u}+5$
$\mathrm{I}=\int_{-5}^{\mathrm{x}-5} f(\mathrm{u}+5) \mathrm{du}$
$=\int_{-5}^{x-5} g(u) d u$
$=\int_{-5}^{x-5} f^{\prime}(u) d u$
$=f(\mathrm{x}-5)-f(-5)$
$=-f(5-\mathrm{x})+f(5)$
$=f(5)-f(5+\mathrm{x})$
$=\int_{5+x}^{5} f^{\prime}(t) d t=\int_{5+x}^{5} g(t) d t$