If ${\int }_0^{100}f\left(x\right)dx=7,$then $\sum _{r=1}^{100}\left({\int }_0^{1}f(r-1+x)dx\right)=$

Solution:

$\begin{array}{l}\sum _{r=1}^{100} \left({\int }_0^1 f(r-1+x)dx\right)\\ ={\int }_0^1 f\left(x\right)dx+{\int }_0^1 f(1+x)dx+{\int }_0^1 f(2+x)dx+\cdots +{\int }_0^1 f(99+x)dx\\ ={\int }_0^1 f\left(x\right)dx+{\int }_1^2 f\left(x\right)dx+{\int }_2^3 f\left(x\right)dx+\cdots +{\int }_{99}^{100} f\left(x\right)dx\\ ={\int }_0^{100} f\left(x\right)dx=7\end{array}$