Question

If $f(x)=\{\begin{array}{ll}0,& \text{ where }x=\frac{n}{n+1},n=1,2,3\dots ..\\ 1,& \text{ else where }\end{array}$, then the value of $\underset{0}{\overset{2}{\int }}f(x)dx$ is -

• A. 1
• B. 0
• C. 2
• D. $\infty$

Answer 1

Answer: (C)

$\underset{0}{\overset{2}{\int }}f(x)dx=\underset{0}{\overset{1/2}{\int }}1\cdot dx+\underset{1/2}{\overset{2/3}{\int }}1\cdot dx+\underset{2/3}{\overset{3/4}{\int }}1\cdot dx+\dots ..+\underset{\frac{n-1}{n}}{\overset{\frac{n}{n+1}}{\int }}1.dx+\dots \dots \dots +\underset{1}{\overset{2}{\int }}1.dx$
$=\left(\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{1}{2}\right)+\left(\frac{3}{4}-\frac{2}{3}\right)+\dots .+\left(\frac{n}{n+1}-\frac{n-1}{n}\right)+\dots .+1=\frac{n}{n+1}+\dots .+1\text{ as }n\to \infty$
taking limit $n\to \infty$
we get $\underset{0}{\overset{2}{\int }}f(x)dx=1+1=2$