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Q.
If $f(x)=\begin{cases} 0, & \text { where } x=\frac{n}{n+1}, n=1,2,3 \ldots . . \\ 1, & \text { else where }\end{cases}$, then the value of $\int\limits_0^2 f(x) d x$ is -
Integrals
Solution:
$ \int\limits_0^2 f(x) d x=\int\limits_0^{1 / 2} 1 \cdot d x+\int\limits_{1 / 2}^{2 / 3} 1 \cdot d x+\int\limits_{2 / 3}^{3 / 4} 1 \cdot d x+\ldots . .+\int\limits_{\frac{n-1}{n}}^{\frac{n}{n+1}} 1 . d x+\ldots \ldots \ldots+\int\limits_1^2 1 . d x $
$ =\left(\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{1}{2}\right)+\left(\frac{3}{4}-\frac{2}{3}\right)+\ldots .+\left(\frac{n}{n+1}-\frac{n-1}{n}\right)+\ldots .+1=\frac{n}{n+1}+\ldots .+1 \text { as } n \rightarrow \infty$
taking limit $n \rightarrow \infty$
we get $\int\limits_0^2 f(x) d x=1+1=2$