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

Let  $I=\underset{0}{\overset{4}{\int }}\frac{1}{1+f\left(x\right)}dx$  …… (1)
$I=\underset{0}{\overset{4}{\int }}\frac{1}{1+f(4-x)}dx,putf(4-x)=\frac{1}{f\left(x\right)}$ 
$I=\underset{0}{\overset{4}{\int }}\frac{f\left(x\right)}{1+f\left(x\right)+1}dx$   ….. (2)
On adding (1) and (2), we get
 $2I=\underset{0}{\overset{4}{\int }}dx\Rightarrow I=2$