If $f(x+f(y))=f(x)+y$ for all $x, y \in R$ and $f(0)=1$, then $\int\limits_0^{10} f(10-x) d x$ is equal to

$10 \int\limits_0^1 f(x) d x$

We have,

$f(x+f(y))=f(x)+y$ for all $x, y \in R$

Replacing $y$ by 0, we obtain

$\Rightarrow f(x+f(0))=f(x)+0$ for all $x \in R$

$\Rightarrow f(x+1)=f(x)$ for all $x \in R$

$\Rightarrow f(x)$ is periodic with period 1 .

Now,

$\int\limits_0^{10} f(10-x) d x=-\int\limits_{10}^0 f(t) d t$, where $t=10-x$

$=\int\limits_0^{10} f(t) d t=10 \int\limits_0^1 f(t) d t$        [∵ f(x) is periodic with period 1]