Question

Let $f:R\to R$ be a continuous function such that $f\left(x\right)+f(x+1)=2$ for all $x\in R.$

If $I_1={\int }_0^{8}f\left(x\right)dxand{I}_2={\int }_{-1}^{3}f\left(x\right)dx$, then the value of $I_1+2I_2$ is equal to:

Answer 1

Determine the value of $I_1+2I_2$:

We have, $f\left(x\right)+f(x+1)=2......\left(1\right)$

Putting $x\to x+1$

$f(x+1)+f(x+2)=2.....\left(2\right)$

Now, equating $\left(1\right)-\left(2\right)$:

$$\begin{gathered} f\left(x\right)+f(x+1)-f(x+2)-f(x+1)=0 \\ \Rightarrow f\left(x\right)=f(x+2) \end{gathered}$$

Therefore, we can say that: $f\left(x\right)=f(x+T)whereT\to period=2$

Solving the integral operation:

$$\begin{gathered} I_1={\int }_0^{8}f\left(x\right)dx \\ \Rightarrow I_1={\int }_0^{8}f\left(x\right)dx \\ \Rightarrow I_1={\int }_0^{2\times 4}f\left(x\right)dx \\ \Rightarrow I_1=4{\int }_0^{2}f\left(x\right)dx \end{gathered}$$

Similarly,

$$\begin{gathered} I_2={\int }_{-1}^{3}f\left(x\right)dx \\ \Rightarrow I_2={\int }_0^{4}f(x+2)dx \\ \Rightarrow I_2={\int }_0^4\left[2-f\left(x\right)\right]dx \\ \Rightarrow I_2=8-2{\int }_0^{2}f\left(x\right)dx \\ \therefore I_1+2I_2=16 \end{gathered}$$

Hence, the value of $I_1+2I_2$ is $16.$