lf f(x) is continuous for all real values of x, then ∑n=110 ∫01 f(r−1+x)dx

# lf f(x) is continuous for all real values of x, then $\sum _{\mathrm{n}=1}^{10} {\int }_{0}^{1} \mathrm{f}\left(\mathrm{r}-1+\mathrm{x}\right)\mathrm{dx}$

1. A

${\int }_{0}^{10} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

2. B

${\int }_{0}^{1} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

3. C

$10{\int }_{0}^{1} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

4. D

$9{\int }_{0}^{1} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

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### Solution:

$\begin{array}{c}\mathrm{I}=\sum _{\mathrm{r}=1}^{10} {\int }_{0}^{1} \mathrm{f}\left(\mathrm{r}-1+\mathrm{x}\right)\mathrm{dx}\\ {\mathrm{I}}_{\mathrm{t}}={\int }_{0}^{1} \mathrm{f}\left(\mathrm{t}-1+\mathrm{x}\right)\mathrm{dx}\end{array}$

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