Thank you for reporting, we will resolve it shortly
Q.
Let $f \left(x\right)= \int\limits_0^{x} g (t) dt$ , where g is a non-zero even function. If $f \left(x+5\right)=g\left(x\right)$, then $ \int\limits_0^{x}f (t) dt $ equals-
$f( x )=\int_{0}^{ x } g ( t ) dt$
$f(- x )=\int_{0}^{- x } g ( t ) dt$
put $t =- u$
$=-\int_{0}^{ x } g (- u ) du$
$=-\int_{0}^{x} g(u) d(u)=-f(x)$
$\Rightarrow f(- x )=-f( x )$
$\Rightarrow f( x )$ is an odd function
Also $f(5+x)=g(x)$
$f(5-x)=g(-x)=g(x)=f(5+x)$
$\Rightarrow f(5-x)=f(5+x)$
Now
$I =\int_{0}^{ x } f( t ) dt$
$t = u +5$
$I =\int_{-5}^{ x -5} f( u +5) du$
$=\int_{-5}^{x-5} g(u) d u$
$=\int_{-5}^{x-5} f^{\prime}(u) d u$
$=f(x-5)-f(-5)$
$=-f(5- x )+f(5)$
$=f(5)-f(5+x)$
$=\int_{5+x}^{5} f^{\prime}(t) d t=\int_{5+x}^{5} g(t) d t$