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Q.
If $\int\limits_0^{100} f(x) d x=a$, then $\displaystyle\sum_{r=1}^{100}\left(\int\limits_0^1 f(r-1+x) d x\right)=$
Integrals
Solution:
$\displaystyle\sum_{r=1}^{100}\left(\int\limits_0^1 f(r-1+x) d x\right) $
$ = \int\limits_0^1 f(x) d x+\int\limits_0^1 f(1+x) d x+\int\limits_0^1 f(2+x) d x+\ldots \ldots+\int\limits_0^1 f(99+x) d x $
$ = \int\limits_0^1 f(x) d x+\int\limits_1^2 f(x) d x+\ldots . .+\int\limits_{99}^{100} f(x) d x $
( using shifting property )
$= \int\limits_0^{100} f(x) d x=a$