# Let $f\left ( x \right )=\int_{0}^{x}g(t)dt,$ where g is a non-zero even function. If $f(x+5)=g(x),$ then $\int_{0}^{x}f(t)dt$ equals : Option 1)$\int_{x+5}^{5}g(t)dt$Option 2)$\int_{5}^{x+5}g(t)dt$Option 3)$2\int_{5}^{x+5}g(t)dt$  Option 4)$5\int_{x+5}^{5}g(t)dt$

$f\left ( x \right )=\int_{0}^{x}g(t)dt$

$f\left ( -x \right )=\int_{0}^{-x}g(t)dt$

put $t=-u$

$=-\int_{0}^{x}g(-u)du=-\int_{0}^{x}g(u)d(u)=-f(x)$

$\because$ g is even function.

$\Rightarrow f\left ( -x \right )=-f(x)$

$\Rightarrow f(x)$ is an odd function

Also given

$\\f(5+x)=g(x)\\f(5-x)=g(-x)=g(x)=f(5+x)$

$\Rightarrow f(5-x)=f(5+x)$

Now, let

$I=\int_{0}^{x}f(x)dt$

$t=u+5$

$I=\int_{-5}^{x-5}f(u+5)du=\int_{-5}^{x-5}g(u)du$

$= \int_{-5}^{x-5}{f}'(u)du\: \: \: \: \: \: \: \: from\cdots (1)$

$=f(x-5)-f(-5)=-f(5-x)+f(5)$

$=f(5)-f(5+x)$

$=\int_{5+x}^{5}{f}'(t)dt=\int_{5+x}^{5}g(t)dt$

Option 1)

$\int_{x+5}^{5}g(t)dt$

Option 2)

$\int_{5}^{x+5}g(t)dt$

Option 3)

$2\int_{5}^{x+5}g(t)dt$

Option 4)

$5\int_{x+5}^{5}g(t)dt$

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