lf f(x) is continuous for all real values of x, then $\sum _{\mathrm{n}=1}^{10} {\int }_0^1 \mathrm{f}(\mathrm{r}-1+\mathrm{x})\mathrm{dx}$

Solution:

$\begin{array}{c}\mathrm{I}=\sum _{\mathrm{r}=1}^{10} {\int }_0^1 \mathrm{f}(\mathrm{r}-1+\mathrm{x})\mathrm{dx}\\ {\mathrm{I}}_{\mathrm{t}}={\int }_0^1 \mathrm{f}(\mathrm{t}-1+\mathrm{x})\mathrm{dx}\end{array}$

$$\begin{gathered} \begin{array}{l}\text{ Put }\mathrm{t}-1+\mathrm{x}=\mathrm{y}\Rightarrow \mathrm{dx}=\mathrm{dy}\\ \begin{array}{ll}{\mathrm{I}}_{\mathrm{t}}={\int }_{\mathrm{t}-1}^{\mathrm{t}} \mathrm{f}\left(\mathrm{y}\right)\mathrm{dy}\Rightarrow & {\mathrm{I}}_{\mathrm{t}}={\int }_{\mathrm{t}-1}^{\mathrm{t}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\\ {\mathrm{I}}_1={\int }_0^1 \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx},{\mathrm{I}}_2={\int }_1^2 \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}& \\ {\mathrm{I}}_3={\int }_2^3 \mathrm{f}\left(\mathrm{x}\right){\mathrm{dx}}_{\mathrm{r}...}& {\mathrm{I}}_{10}={\int }_9^{10} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\end{array}\end{array} \\ I = I_1+I_2+....+I_{10}={\int }_0^1 \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_1^2 \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+........+{\int }_9^{10} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{10} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \end{gathered}$$