Question

The value of ${\int }_0^{2}f\left(x\right)dx;$ where $f\left(x\right)=\{\begin{array}{c}0,whenx=\frac{n}{n+1},n=1,2,3,...\\ 1,elsewhere\end{array}$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $noneofthese$

Answer 1

Answer: (A)

$1$

${\int }_0^{2}f\left(x\right)dx$

$={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$

$={\int }_0^{1}0dx+{\int }_1^{2}dx$

$=0+{\left[x\right]}_1^2=(2-1)=1$