JEE Main 2021 (Online) 16th March Morning ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be a continuous function such that $f(x)+f(x+1)=2$, for all $x\in\mathbf{R}$. If $I_1=\displaystyle\int_{0}^{8}f(x)\,dx$ and $I_2=\displaystyle\int_{-1}^{3}f(x)\,dx$, then the value of $I_1+2I_2$ is equal to $\_\_\_\_\_\_\_\_$ .
Definite Integration
Answer$16$
Solution
Concept. $f(x)+f(x+1)=2$ means any two consecutive unit intervals together give $2$. Formulas used. (1) $\int_k^{k+2}f=\int_k^{k+1}\big(f(x)+f(x+1)\big)dx=2$Step 1 - $I_1$. $[0,8]$ is four blocks of length $2$. By (1), $I_1=8$. Step 2 - $I_2$. $[-1,3]$ is two blocks of length $2$, so $I_2=4$. Step 3 - Value. $I_1+2I_2=8+8=16$. Answer: 16