Question

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

Answer 1

Hint: For solving this question first we will do some substitutions like replace $x\to -x$ in $f\left( x+5 \right)=g\left( x \right)$ and solve further to prove that $f$ is an odd function. After that, we will differentiate $f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt$ with respect to $x$ for proving \[{f}'\left( x \right)=g\left( x \right)\] and further we will substitute $t=u+5$ in the integral $\int\limits_{0}^{x}{f\left( t \right)}dt$ and solve for the final answer.

Complete step-by-step solution -
Given:
It is given that $f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt$ , where $g$ is a non-zero even function and $f\left( x+5 \right)=g\left( x \right)$ , then we have to solve for $\int\limits_{0}^{x}{f\left( t \right)}dt$ .
Now, it is given that $g$ is a non-zero even function. Then,
$g\left( x \right)=g\left( -x \right)...................\left( 1 \right)$
Now, as it is given that $f\left( x+5 \right)=g\left( x \right)$ and if we replace $x\to -x$ . Then,
$\begin{align}
  & f\left( x+5 \right)=g\left( x \right) \\
 & \Rightarrow f\left( 5-x \right)=g\left( -x \right) \\
\end{align}$
Now, from equation (1) we can write $g\left( x \right)=g\left( -x \right)$ . Then,
$\begin{align}
  & f\left( 5-x \right)=g\left( -x \right) \\
 & \Rightarrow f\left( 5-x \right)=g\left( x \right) \\
\end{align}$
Now, we know that $f\left( x+5 \right)=g\left( x \right)$ . Then,
$\begin{align}
  & f\left( 5-x \right)=g\left( x \right) \\
 & \Rightarrow f\left( 5-x \right)=f\left( x+5 \right)................\left( 2 \right) \\
\end{align}$
Now, as it is given that $f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt$ and if we replace $x\to -x$ . Then,
$\begin{align}
  & f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt \\
 & \Rightarrow f\left( -x \right)=\int\limits_{0}^{-x}{g\left( t \right)}dt \\
\end{align}$
Now, substitute $t=-u$ in the above integral then, the differential $dt$ will change accordingly, upper and lower limits of the integral $\int\limits_{0}^{-x}{g\left( t \right)}dt$ will change. And for the upper limit, we will find $u$ when $t=-x$ similarly, for the lower limit find $u$ when $t=0$ . Then,
$\begin{align}
  & t=-u \\
 & \Rightarrow t=0=-u \\
 & \Rightarrow u=0 \\
 & t=-u \\
 & \Rightarrow t=-x=-u \\
 & \Rightarrow u=x \\
 & t=-u \\
 & \Rightarrow dt=-du \\
 & f\left( -x \right)=\int\limits_{0}^{-x}{g\left( t \right)}dt \\
 & \Rightarrow f\left( -x \right)=-\int\limits_{0}^{x}{g\left( -u \right)}du \\
\end{align}$
Now, as it is given that $g$ is a non-zero even function, we can write $g\left( -u \right)=g\left( u \right)$. Then,
$\begin{align}
  & f\left( -x \right)=-\int\limits_{0}^{x}{g\left( -u \right)}du \\
 & \Rightarrow f\left( -x \right)=-\int\limits_{0}^{x}{g\left( u \right)}du \\
\end{align}$
Now, as it is given that $f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt$ so, we can write $\int\limits_{0}^{x}{g\left( u \right)}du=f\left( x \right)$ in the above equation. Then,
$\begin{align}
  & f\left( -x \right)=-\int\limits_{0}^{x}{g\left( u \right)}du \\
 & \Rightarrow f\left( -x \right)=-f\left( x \right)....................\left( 3 \right) \\
\end{align}$
Now, from the above result, we conclude that $f$ is an odd function.
Now, we will differentiate the $f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt$ with respect to $x$ . Then,
\[\begin{align}
  & f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt \\
 & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( \int\limits_{0}^{x}{g\left( t \right)}dt \right)}{dx} \\
 & \Rightarrow {f}'\left( x \right)=g\left( x \right)...........................\left( 4 \right) \\
\end{align}\]
Now, as it is given that $f\left( x+5 \right)=g\left( x \right)$ . Then,
\[{f}'\left( x \right)=f\left( x+5 \right)..................\left( 5 \right)\]
Now, we will solve for $\int\limits_{0}^{x}{f\left( t \right)}dt$ . First substitute $t=u+5$ then, the differential $dt$ will change accordingly, upper and lower limits of the integral $\int\limits_{0}^{x}{f\left( t \right)}dt$ will change. And for the upper limit, we will find $u$ when $t=x$ similarly, for the lower limit find $u$ when $t=0$ . Then,

$\begin{align}
  & t=u+5 \\
 & \Rightarrow t=0=u+5 \\
 & \Rightarrow u=-5 \\
 & t=u+5 \\
 & \Rightarrow t=x=u+5 \\
 & \Rightarrow u=x-5 \\
 & t=u+5 \\
 & \Rightarrow dt=du \\
 & \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{-5}^{x-5}{f\left( u+5 \right)}du \\
\end{align}$
Now, from equation (5) we know that \[{f}'\left( x \right)=f\left( x+5 \right)\] so, we can replace $f\left( u+5 \right)={f}'\left( u \right)$ . Then,
$\begin{align}
  & \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{-5}^{x-5}{f\left( u+5 \right)}du \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{-5}^{x-5}{{f}'\left( u \right)}du \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=\left[ f\left( u \right) \right]_{-5}^{x-5} \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( x-5 \right)-f\left( -5 \right) \\
\end{align}$
Now, from equation (3) we know that $f\left( -x \right)=-f\left( x \right)$ so, we can write $f\left( x-5 \right)=-f\left( 5-x \right)$ and $f\left( -5 \right)=-f\left( 5 \right)$ in the above equation. Then,
$\begin{align}
  & \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( x-5 \right)-f\left( -5 \right) \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=-f\left( 5-x \right)+f\left( 5 \right) \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( 5 \right)-f\left( 5-x \right) \\
\end{align}$
Now, from equation (2) we can write $f\left( 5-x \right)=f\left( x+5 \right)$ in the above equation. Then,
\[\begin{align}
  & \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( 5 \right)-f\left( 5-x \right) \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( 5 \right)-f\left( x+5 \right) \\
\end{align}\]
Now, we can write \[f\left( 5 \right)-f\left( x+5 \right)=\int\limits_{5+x}^{5}{{f}'\left( t \right)dt}\] in the above equation. Then,
\[\begin{align}
  & \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( 5 \right)-f\left( x+5 \right) \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{5+x}^{5}{{f}'\left( t \right)dt} \\
\end{align}\]
Now, from equation (4) we know that \[{f}'\left( x \right)=g\left( x \right)\] so, we can write \[\int\limits_{5+x}^{5}{{f}'\left( t \right)dt}=\int\limits_{5+x}^{5}{g\left( t \right)dt}\] in the above equation. Then,
\[\begin{align}
  & \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{5+x}^{5}{{f}'\left( t \right)dt} \\
 & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{5+x}^{5}{g\left( t \right)dt} \\
\end{align}\]
Now, from the above result, we conclude that \[\int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{5+x}^{5}{g\left( t \right)dt}\] .
Hence, (a) will be the correct option.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer. Moreover, we should proceed in a stepwise manner for smooth calculation and make the right substitutions while solving the question. And we should avoid making calculation mistakes while solving.