Question

If $f\left(x\right)$ is continuous for all real values of x, then $\sum _{r=1}^{10}{\int }_0^{1}f(r-1+x)dx$ is equal to.

• A. ${\int }_0^{0}f\left(x\right)dx$
• B. ${\int }_0^{1}f\left(x\right)dx$
• C. $10{\int }_0^{1}f\left(x\right)dx$
• D. $9{\int }_0^{1}f\left(x\right)dx$

Answer 1

The correct option is A ${\int }_0^{0}f\left(x\right)dx$

Given : ${\sum }_{r=1}^{10}{\int }_0^{1}f(r-1+x)dx$

${\sum }_{r=1}^{10}{\int }_0^{1}f(r-1+x)dx={\sum }_{r=1}^{10}{\int }_0^{1}f(r-x)dxLetr-x=tx\to 0t\to r-dx=dtx\to 1t\to r-1{\sum }_{r=1}^{10}{\int }_0^{1}f(r-x)dx={\sum }_{r=1}^{10}{\int }_{r-1}^{r}f\left(t\right)dx={\sum }_{r=1}^{10}\left[f\right(r)-f(r-1\left)\right]={\sum }_{r=1}^{10}f\left(r\right)-{\sum }_{r=1}^{10}f(r-1)=f\left(10\right)-f\left(0\right)={\int }_0^{10}f\left(t\right)dt$

Hence the correct answer is ${\int }_0^{10}f\left(t\right)dt$