Question

Let $f\left(x\right)$ be a continuous function on $[0,4]$ satisfying $f\left(x\right)f(4-x)=1$
The value of the definite integral ${\int }_0^4\frac{1}{1+f\left(x\right)}dx$ equals

• A. $0$
• B. $4$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$I={\int }_0^4\frac{1}{1+f\left(x\right)}dx={\int }_0^4\frac{1}{1+f(4-x)}dx$

$={\int }_0^4\frac{1}{1+\frac{1}{f\left(x\right)}}dx$

$={\int }_0^4\frac{f\left(x\right)}{f\left(x\right)+1}dx$------(ii)

Add (1) & (ii)

$2I={\int }_0^{4}dx=4$

$\Rightarrow I=2$