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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

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

$\int\limits_0^8 {f(x)dx + 2\int\limits_{ - 1}^3 {f(x)dx} } = ?$

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 $4\int\limits_0^2 {f(x)dx}  - 4\int\limits_0^2 {f(x)dx}  + 16 = 16$