Question

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

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

Answer 1

The correct option is A $\underset{x+5}{\overset{5}{\int }}g\left(t\right)dt$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}g\left(t\right)dt$
$\Rightarrow f^{\prime }\left(x\right)=g\left(x\right)$
Given, $g$ is even $\Rightarrow f^{\prime }$ is even.
$\Rightarrow f$ is odd.

$f(x+5)=g\left(x\right)$
$\Rightarrow f\left(x\right)=g(x-5)\cdots \left(1\right)$
Now, consider $I=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
Put $t=u-5\Rightarrow dt=du$
$I=\underset{5}{\overset{x+5}{\int }}f(u-5)du$
$\Rightarrow I=\underset{5}{\overset{x+5}{\int }}f(-(5-u\left)\right)du$
$\Rightarrow I=-\underset{5}{\overset{x+5}{\int }}f(5-u)du$
$\Rightarrow I=-\underset{5}{\overset{x+5}{\int }}g(-u)du\left[\text{From}\right(1\left)\right]$
$\Rightarrow I=-\underset{5}{\overset{x+5}{\int }}g\left(u\right)du$
$\Rightarrow I=\underset{x+5}{\overset{5}{\int }}g\left(t\right)dt$