Question

If $f\left(x\right)=\left\{\begin{array}{cc}0,& wherex=\frac{n}{n+1},n=1,2,\mathrm{3.....},\\ 1,& elsewhere\end{array}\right\}$ then the value of ${\int }_0^{2}f\left(x\right)dx$

• A. $1$
• B. $0$
• C. $2$
• D. $\infty$

Answer 1

Answer: (A)

$1$

Given : $f\left(x\right)=\{0x=\frac{n}{n+1}(n=1,2,\mathrm{3....}\left)\right\}$

$\left\{1elsewhere\right\}$

We find that for $x\in (0,1)f\left(x\right)=0$

${\int }_0^{2}f\left(x\right)dx={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx={\int }_0^{1}0dx+{\int }_1^{2}1dx={\left[x\right]}_1^2=2-1=1$

Hence the correct answer is $1$