Let $f:R \to R$ be a continuous function such that $f(x) + f(x + 1) = 2$, for all $x\in R$. If ${I_1} = \int\limits_0^8 {f(x)dx} $ and ${I_2} = \int\limits_{ – 1}^3 {f(x)dx} $, then the value of $I_1 +2I_2 $ is equal to ……

Solution

$x \to t + 1$ in $I_2$ we have ${I_2} = \int\limits_{ – 2}^2 {f(t + 1)dt} $

${I_1} + 2{I_2} = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ – 2}^2 {f(t + 1)dt} $

$= \int\limits_0^8 {f(x)dx} + 2\int\limits_{ – 2}^2 {f(x + 1)dx} $

$ = \int\limits_0^8 {f(x)dx} + 2\int\limits_{ – 2}^2 {[2 – f(x)]dx} $

$ = \int\limits_0^8 {f(x)dx} – 2\int\limits_{ – 2}^2 {f(x)dx} + 4\int\limits_{ – 2}^2 {dx} $

$ = \int\limits_0^8 {f(x)dx} – 2\int\limits_{ – 2}^2 {f(x)dx} + 16$

$x \to x + 1$ in $f(x) + f(x + 1) = 2$ yields $f(x + 1) + f(x + 2) = 2$

$\therefore f(x) + f(x + 1) = f(x + 1) + f(x + 2)$

$\Rightarrow f(x+2)=f(x)$ which means the function is periodic with period 2.

So, $I_1 +2I_2$ reduces to $4\int\limits_0^2 {f(x)dx} – 4\int\limits_0^2 {f(x)dx} + 16 = 16$