Question

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

• 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 ${\int }_0^{n}f\left(x\right)dx$

${\sum }_{r=1}^n{\int }_0^{1}f(r-1+x)dx={\int }_0^{1}f\left(x\right)dx+{\int }_0^{1}f(1+x)dx+{\int }_0^{1}f(2+x)dx+...+{\int }_0^{1}f(n-1+x)dx={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx+{\int }_2^{3}f\left(x\right)dx+...+{\int }_{n=1}^{n}f\left(x\right)dx={\int }_0^{n}f\left(x\right)dx$