A) \[1\]
B) \[2\]
C) \[0\]
D) \[1/2\]
Correct Answer: A
Solution :
Given, \[f(x)=x-[x]\] \[\therefore \] \[\int_{-1}^{1}{f(x)}dx=\int_{-1}^{1}{(x-[x])dx}\] \[=\int_{-1}^{0}{(x-[x])dx+\int_{0}^{1}{(x-[x])dx}}\] \[=\int_{-1}^{0}{(x+1)dx+\int_{1}^{0}{x}\,\,dx}\] \[=\left[ \frac{{{x}^{2}}}{2}+x \right]_{-1}^{0}+\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}\] \[=\left[ 0-\frac{1}{2}+1 \right]+\left[ \frac{1}{2}-0 \right]\] \[=\frac{1}{2}+\frac{1}{2}=1\] Note: For every real value of x, the value of \[[x]\] is always an integers.
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