Q66 / 120 · VITEEE 2013 · Mathematics
If $f(x)=x-[x]$, for every real number $x$, where $[x]$, is the integral part of $x$. Then, $\int_{-1}^{1} f(x) d x$ is equal to
- Correct answerA1
- B2
- C0
- D$\frac{1}{2}$
Solution
$\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}],-1 \leq \mathrm{x} < 0$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}+1$
When $0 \leq \mathrm{x} < 1$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}$
$\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{-1}^{0} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
$=\int_{-1}^{0}(\mathrm{x}+1) \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{x} \mathrm{dx}$
$=\left[\frac{\mathrm{x}^{2}}{2}+\mathrm{x}\right]_{-1}^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}$
$=0-\left[\frac{(-1)^{2}}{2}-1\right]+\frac{1}{2}=1$
Hence, the correct answer is option: 'A'.