Question

If $f(x)=f(a+x)$ and $\int_{0}^{a}\:f(x)\:dx\:=p$ then, find the value of
$\int^{na}_{a} \: f(x) \: dx$
$\rm \: (A) \: np \: \: \: \: (B)(n - 1)p \: \: \: \: (C)(n + 1)p \: \: \: \: (D)none \: \: of \: \: these$
​

Answer 1

Given,

$\small\longrightarrow f(x)=f(a+x)$

It means |a| can be period of f(x).

$\small\Longrightarrow f(x)=f(\lambda a+x)\quad\forall\lambda\in\mathbb{Z}\quad\dots(1)$

Then it is true that,

$\small\displaystyle\longrightarrow\int\limits_m^{m+a}f(x)\,dx=\int\limits_0^af(x)\,dx\quad\dots(2)$

where m is a real number such that m ∈ [(k - 1)a, ka) for some k ∈ Z.

Proof is given below.

$\small\displaystyle\longrightarrow\int\limits_m^{m+a}f(x)\,dx=\int\limits_m^{ka}f(x)\,dx+\int\limits_{ka}^{m+a}f(x)\,dx$

We know,

• $\small\displaystyle\int\limits_p^qf(x)\,dx=\int\limits_{p-r}^{q-r}f(x+r)\,dx\quad\dots(3)$

Then,

$\small\displaystyle\longrightarrow\int\limits_m^{m+a}f(x)\,dx=\int\limits_{m-(k-1)a}^{ka-(k-1)a}f(x+(k-1)a)\,dx+\int\limits_{ka-ka}^{m+a-ka}f(x+ka)\,dx$

$\small\displaystyle\longrightarrow\int\limits_m^{m+a}f(x)\,dx=\int\limits_{m+a-ka}^{a}f((k-1)a+x)\,dx+\int\limits_{0}^{m+a-ka}f(ka+x)\,dx$

From (1),

$\small\displaystyle\longrightarrow\int\limits_m^{m+a}f(x)\,dx=\int\limits_{m+a-ka}^{a}f(x)\,dx+\int\limits_{0}^{m+a-ka}f(x)\,dx$

$\small\displaystyle\longrightarrow\int\limits_m^{m+a}f(x)\,dx=\int\limits_0^af(x)\,dx$

Taking m = (k - 1)a,

$\small\displaystyle\longrightarrow\int\limits_{(k-1)a}^{(k-1)a+a}f(x)\,dx=\int\limits_0^af(x)\,dx$

$\small\displaystyle\longrightarrow\int\limits_{(k-1)a}^{ka}f(x)\,dx=\int\limits_0^af(x)\,dx\quad\dots(4)$

Hence,

$\small\text{ \displaystyle\longrightarrow\int\limits_a^{na}f(x)\,dx=\underbrace{\int\limits_a^{2a}f(x)\,dx+\int\limits_{2a}^{3a}f(x)\,dx+\int\limits_{3a}^{4a}f(x)\,dx+\,\dots\,+\int\limits_{(n-1)a}^{na}f(x)\,dx}_{\frac{na-2a}{a}+1=(n-1)\,terms} }$

[No. of integrals in RHS (n - 1) is got by considering each upper limit as in AP. One can consider lower limit also instead of upper limit.]

By (4) each integral in RHS becomes,

$\small\displaystyle\longrightarrow\int\limits_a^{na}f(x)\,dx=\underbrace{\int\limits_0^af(x)\,dx+\int\limits_0^af(x)\,dx+\int\limits_0^af(x)\,dx+\,\dots\,+\int\limits_0^af(x)\,dx}_{(n-1)\,terms}$

$\small\displaystyle\longrightarrow\int\limits_a^{na}f(x)\,dx=(n-1)\int\limits_0^af(x)\,dx$

Given,

$\small\displaystyle\longrightarrow\int\limits_0^af(x)\,dx=p$

Then,

$\small\text{ \displaystyle\longrightarrow\underline{\underline{\int\limits_a^{na}f(x)\,dx=(n-1)p}} }$

Hence (B) is the answer.