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Question: Let \(f\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt\) , where \(g\) is a non-zero even ...

Let f(x)=∫0xg(t)dtf\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt , where gg is a non-zero even function. If f(x+5)=g(x)f\left( x+5 \right)=g\left( x \right) , then ∫0xf(t)dt\int\limits_{0}^{x}{f\left( t \right)}dt equals.
(a) ∫x+55g(t)dt\int\limits_{x+5}^{5}{g\left( t \right)}dt
(b) 5∫x+55g(t)dt5\int\limits_{x+5}^{5}{g\left( t \right)}dt
(c) ∫5x+5g(t)dt\int\limits_{5}^{x+5}{g\left( t \right)}dt
(d) 5∫5x+5g(t)dt5\int\limits_{5}^{x+5}{g\left( t \right)}dt

Explanation

Solution

Hint: For solving this question first we will do some substitutions like replace x→−xx\to -x in f(x+5)=g(x)f\left( x+5 \right)=g\left( x \right) and solve further to prove that ff is an odd function. After that, we will differentiate f(x)=∫0xg(t)dtf\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt with respect to xx for proving f′(x)=g(x){f}'\left( x \right)=g\left( x \right) and further we will substitute t=u+5t=u+5 in the integral ∫0xf(t)dt\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(x)=∫0xg(t)dtf\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt , where gg is a non-zero even function and f(x+5)=g(x)f\left( x+5 \right)=g\left( x \right) , then we have to solve for ∫0xf(t)dt\int\limits_{0}^{x}{f\left( t \right)}dt .
Now, it is given that gg is a non-zero even function. Then,
g(x)=g(−x)...................(1)g\left( x \right)=g\left( -x \right)...................\left( 1 \right)
Now, as it is given that f(x+5)=g(x)f\left( x+5 \right)=g\left( x \right) and if we replace x→−xx\to -x . Then,
f(x+5)=g(x) ⇒f(5−x)=g(−x) \begin{aligned} & f\left( x+5 \right)=g\left( x \right) \\\ & \Rightarrow f\left( 5-x \right)=g\left( -x \right) \\\ \end{aligned}
Now, from equation (1) we can write g(x)=g(−x)g\left( x \right)=g\left( -x \right) . Then,
f(5−x)=g(−x) ⇒f(5−x)=g(x) \begin{aligned} & f\left( 5-x \right)=g\left( -x \right) \\\ & \Rightarrow f\left( 5-x \right)=g\left( x \right) \\\ \end{aligned}
Now, we know that f(x+5)=g(x)f\left( x+5 \right)=g\left( x \right) . Then,
f(5−x)=g(x) ⇒f(5−x)=f(x+5)................(2) \begin{aligned} & f\left( 5-x \right)=g\left( x \right) \\\ & \Rightarrow f\left( 5-x \right)=f\left( x+5 \right)................\left( 2 \right) \\\ \end{aligned}
Now, as it is given that f(x)=∫0xg(t)dtf\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt and if we replace x→−xx\to -x . Then,
f(x)=∫0xg(t)dt ⇒f(−x)=∫0−xg(t)dt \begin{aligned} & 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{aligned}
Now, substitute t=−ut=-u in the above integral then, the differential dtdt will change accordingly, upper and lower limits of the integral ∫0−xg(t)dt\int\limits_{0}^{-x}{g\left( t \right)}dt will change. And for the upper limit, we will find uu when t=−xt=-x similarly, for the lower limit find uu when t=0t=0 . Then,
t=−u ⇒t=0=−u ⇒u=0 t=−u ⇒t=−x=−u ⇒u=x t=−u ⇒dt=−du f(−x)=∫0−xg(t)dt ⇒f(−x)=−∫0xg(−u)du \begin{aligned} & 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{aligned}
Now, as it is given that gg is a non-zero even function, we can write g(−u)=g(u)g\left( -u \right)=g\left( u \right). Then,
f(−x)=−∫0xg(−u)du ⇒f(−x)=−∫0xg(u)du \begin{aligned} & 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{aligned}
Now, as it is given that f(x)=∫0xg(t)dtf\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt so, we can write ∫0xg(u)du=f(x)\int\limits_{0}^{x}{g\left( u \right)}du=f\left( x \right) in the above equation. Then,
f(−x)=−∫0xg(u)du ⇒f(−x)=−f(x)....................(3) \begin{aligned} & 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{aligned}
Now, from the above result, we conclude that ff is an odd function.
Now, we will differentiate the f(x)=∫0xg(t)dtf\left( x \right)=\int\limits_{0}^{x}{g\left( t \right)}dt with respect to xx . Then,

& 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{aligned}$$ Now, as it is given that $f\left( x+5 \right)=g\left( x \right)$ . Then, $${f}'\left( x \right)=f\left( x+5 \right)..................\left( 5 \right)$$ Now, we will solve for $\int\limits_{0}^{x}{f\left( t \right)}dt$ . First substitute $t=u+5$ then, the differential $dt$ will change accordingly, upper and lower limits of the integral $\int\limits_{0}^{x}{f\left( t \right)}dt$ will change. And for the upper limit, we will find $u$ when $t=x$ similarly, for the lower limit find $u$ when $t=0$ . Then, $\begin{aligned} & t=u+5 \\\ & \Rightarrow t=0=u+5 \\\ & \Rightarrow u=-5 \\\ & t=u+5 \\\ & \Rightarrow t=x=u+5 \\\ & \Rightarrow u=x-5 \\\ & t=u+5 \\\ & \Rightarrow dt=du \\\ & \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{-5}^{x-5}{f\left( u+5 \right)}du \\\ \end{aligned}$ Now, from equation (5) we know that $${f}'\left( x \right)=f\left( x+5 \right)$$ so, we can replace $f\left( u+5 \right)={f}'\left( u \right)$ . Then, $\begin{aligned} & \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{-5}^{x-5}{f\left( u+5 \right)}du \\\ & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=\int\limits_{-5}^{x-5}{{f}'\left( u \right)}du \\\ & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=\left[ f\left( u \right) \right]_{-5}^{x-5} \\\ & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( x-5 \right)-f\left( -5 \right) \\\ \end{aligned}$ Now, from equation (3) we know that $f\left( -x \right)=-f\left( x \right)$ so, we can write $f\left( x-5 \right)=-f\left( 5-x \right)$ and $f\left( -5 \right)=-f\left( 5 \right)$ in the above equation. Then, $\begin{aligned} & \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( x-5 \right)-f\left( -5 \right) \\\ & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=-f\left( 5-x \right)+f\left( 5 \right) \\\ & \Rightarrow \int\limits_{0}^{x}{f\left( t \right)}dt=f\left( 5 \right)-f\left( 5-x \right) \\\ \end{aligned}$ Now, from equation (2) we can write $f\left( 5-x \right)=f\left( x+5 \right)$ in the above equation. Then, $$\begin{aligned} & \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{aligned}$$ Now, we can write $$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{aligned} & \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{aligned}$$ Now, from equation (4) we know that $${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{aligned} & \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{aligned}$$ Now, from the above result, we conclude that $$\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.