Question

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

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

Answer 1

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

$I=\sum _{r=1}^n{\int }_0^{1}f(r-1+x)dx$
Substitute $r-1+x=t\Rightarrow dx=dt$ and limit of integral changes from ${\int }_0^1$ to ${\int }_{r-1}^r$
$I=\sum _{r=1}^n{\int }_0^{1}f(r-1+x)dx=\sum _{r=1}^n{\int }_{r-1}^{r}f\left(t\right)dt$
$I={\int }_0^{1}f\left(t\right)dt+{\int }_1^{2}f\left(t\right)dt+\cdots +{\int }_{n-1}^{n}f\left(t\right)dt$
$I={\int }_0^{n}f\left(t\right)dt\equiv {\int }_0^{n}f\left(x\right)dx$