Question

The value of $\sum _{r=1}^n\underset{0}{\overset{1}{\int }}f(r-1+x)dx$ is equal to (if function has period 1)

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

Answer 1

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

${\int }_0^{1}f(r-1+x)dx$

$=\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(10)+f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+.....+f\left(\frac{n-1}{n}\right)]$

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{2=0}^{n=1}f\left(\frac{2}{n}\right)$

$={\int }_0^{n}f\left(x\right)dx$.

Option D is correct.