Definite Integration Question 31
Question 31
- 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 $0 \leq x<1$
$$ 1 \text {, if } \quad x=1 $$
$\therefore \quad \int_{-1}^{1}[x] d x=\int_{-1}^{0}[x] d x+\int_{0}^{1}[x] d x$ $=\int_{-1}^{0}(-1) d x+\int_{0}^{1} 0 d x$ $=-[x]{-1}^{0}+0=-1 ; \quad \therefore \quad \int{-1}^{1} f(x) d x=1$
