Prove the following identities.

Solution:

$f(x)=\int_{0}^{x} g(t) d t$

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

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

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

$\mathrm{t}=\mathrm{u}+5$

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

$=\int_{-5}^{x-5} g(u) d u$

$=\int_{-5}^{x-5} f^{\prime}(u) d u$

$=f(\mathrm{x}-5)-f(-5)$

$=-f(5-\mathrm{x})+f(5)$

$=f(5)-f(5+\mathrm{x})$

$=\int_{5+x}^{5} f^{\prime}(t) d t=\int_{5+x}^{5} g(t) d t$