Question

Let $f\left(x\right)=\underset{0}{\overset{x}{\int }}g(t)dt$ , where g is a non-zero even function. If $f\left(x+5\right)=g\left(x\right)$, then $\underset{0}{\overset{x}{\int }}f(t)dt$ equals-

• A. $\underset{x+5}{\overset{5}{\int }}g(t)dt$
• B. $5\underset{x+5}{\overset{5}{\int }}g(t)dt$
• C. $\underset{5}{\overset{x+5}{\int }}g(t)dt$
• D. $2\underset{5}{\overset{x+5}{\int }}g(t)dt$

Answer 1

Answer: (A)

$f(x)={\int }_0^{x}g(t)dt$
$f(-x)={\int }_0^{-x}g(t)dt$
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
$I={\int }_0^{x}f(t)dt$
$t=u+5$
$I={\int }_{-5}^{x-5}f(u+5)du$
$={\int }_{-5}^{x-5}g(u)du$
$={\int }_{-5}^{x-5}{f}^{\prime }(u)du$
$=f(x-5)-f(-5)$
$=-f(5-x)+f(5)$
$=f(5)-f(5+x)$
$={\int }_{5+x}^5{f}^{\prime }(t)dt={\int }_{5+x}^{5}g(t)dt$