Indefinite Integration 1 Question 31
32. Let $f(x)=x-[x]$, for every real number $x$, where $[x]$ is the integral part of $x$. Then, $\int _{-1}^{1} f(x) d x$ is
(1998, 2M)
(a) 1
(b) 2
(c) 0
(d) $-\frac{1}{2}$
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Solution:
- Let $\int _{-1}^{1} f(x) d x=\int _{-1}^{1}(x-[x]) d x=\int _{-1}^{1} x d x-\int _{-1}^{1}[x] d x$ $=0-\int _{-1}^{1}[x] d x \quad[\because x$ is an odd function $]$
$$ -1, \text { if }-1 \leq x<0 $$
But
$[x]=0$, if $0 \leq x<1$
$$ 1 \text {, if } \quad x=1 $$
$\therefore \quad \int _{-1}^{1}[x] d x=\int _{-1}^{0}[x] d x+\int _0^{1}[x] d x$ $=\int _{-1}^{0}(-1) d x+\int _0^{1} 0 d x$ $=-[x] _{-1}^{0}+0=-1 ; \quad \therefore \quad \int _{-1}^{1} f(x) d x=1$
