Indian Statistical Institute B.Math & B.Stat : Integration

Let $n$ be a positive integer Define $$f(x) = min \{|x-1|,|x-2|,\dots,|x-n|\}$$ Then evaluate $$\int_{0}^{n+1} f(x) dx$$ When $0 < x < 1+ \frac{1}{2}$, $|x-1|=min \{|x-1|,|x-2|,\dots,|x-n|\}$. Now $|x-1|= 1-x$ for $0 < x < 1$ and $|x-1|= x-1$ for $1 < x < 1+ \frac{1}{2}$ So, $\int_{0}^{1+\frac{1}{2}} f(x) dx = \int_{0}^{1} (1-x) dx + \int_{1}^{1+ \frac{1}{2}} (x-1) dx = \frac{1}{2}+ \frac{1}{2} \times \frac{1}{4} \dots (A)$ When $n+ \frac{1}{2} < x < n+1$, $|x-n|=min \{|x-1|,|x-2|,\dots,|x-n|\}$. Now $|x-n|= x-n$ for $n+ \frac{1}{2} < x < n+1$ So, $\int_{n+\frac{1}{2}}^{n+1} f(x) dx = \int_{n+\frac{1}{2}}^{n+1} (x-n) dx = \frac{1}{2}- \frac{1}{2} \times \frac{1}{4} \dots (B)$ Consider the diagram given below where $1 < k \leq n$,. When $x \in \big(k-\frac{1}{2},k+\frac{1}{2} \big)$, $|x-k|$ is minimum among $|x-i|$ where $i=1,2,3,\dots,k-1,k+1,\dots,n$ So, $\int_{k-\frac{1}{2}}^{k+\frac{1}{2}} f(x) dx = \int_{k-\frac{1}{2}}^{k+\frac{1}{2}} |x-k| dx = \int_{k-\frac{1}{2}}^{k} (k-x) dx + \int_{k}^{k+\frac{1}{2}} (x-k) dx = \frac{1}{4}$ for $k=2,3,4\dots,n$. Summing for $k=2,3,4\dots,n$ we get $$\int_{1+\frac{1}{2}}^{n+\frac{1}{2}} f(x) dx = \frac{(n-1)}{4} \dots (C)$$ Adding equations $A,C$ and $B$ we get $$\int_{0}^{n+1} f(x) dx = \int_{0}^{1+\frac{1}{2}} f(x) dx+ \int_{1+\frac{1}{2}}^{n+\frac{1}{2}} f(x) dx+\int_{n+\frac{1}{2}}^{n+1} f(x) dx = \frac{1}{2}+ \frac{1}{2} \times \frac{1}{4}+ \frac{(n-1)}{4}+ \frac{1}{2}- \frac{1}{2} \times \frac{1}{4}$$ $$= \frac{(n-1)}{4}+1 = \frac{n+3}{4}$$