Question

If $f\left(x\right)$ is continuous for all real values of $x$, then $\sum _{r=1}^n\underset{0}{\overset{1}{\int }}f(r-1+x)dx$ is equal to,where n is a natural number

• A. $\underset{0}{\overset{n}{\int }}f\left(x\right)dx$
• B. $\underset{0}{\overset{1}{\int }}f\left(x\right)dx$
• C. $n\underset{0}{\overset{1}{\int }}f\left(x\right)dx$
• D. $(n-1)\underset{0}{\overset{1}{\int }}f\left(x\right)dx$

Answer 1

The correct option is A $\underset{0}{\overset{n}{\int }}f\left(x\right)dx$

$\sum _{r=1}^n\underset{0}{\overset{1}{\int }}f(r-1+x)dx$
$=\underset{0}{\overset{1}{\int }}f\left(x\right)dx+\underset{0}{\overset{1}{\int }}f(1+x)dx+\underset{0}{\overset{1}{\int }}f(2+x)dx+....+\underset{0}{\overset{1}{\int }}f(n-1+x)dx$
$=\underset{0}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{2}{\int }}f\left(x\right)dx+\underset{2}{\overset{3}{\int }}f\left(x\right)dx+....\underset{r-1}{\overset{r}{\int }}f\left(x\right)dx+....+\underset{n-1}{\overset{n}{\int }}f\left(x\right)dx={\int }_0^{n}f\left(x\right)dx$