If f(x) is continuous for all real values of x, then $\sum\limits_{r=1}^n\int\limits_0^1f(r-1+x)dx$ is equal to:
Answer & explanation
Correct answer: option 1
$S=\int\limits_0^1[f(x)+f(1+x)+f(2+x)+......+f(n-1+x)]dx$
$=\int\limits_0^1f(x)dx+\int\limits_1^2f(x)dx+\int\limits_2^3f(x)dx+.....+\int\limits_{n-1}^nf(x)dx=\int\limits_0^nf(x)dx$