A) 1
B) 2
C) 3
D) 0
E) \[\frac{1}{2}\]
Correct Answer: A
Solution :
\[\therefore \] \[\int\limits_{-1}^{1}{(x-[x])dx=}\int\limits_{-1}^{0}{(x-[x])}dx\] \[+\int\limits_{0}^{1}{(x-[x])}dx\] \[=\int\limits_{-1}^{0}{(x+1)}dx+\int\limits_{0}^{1}{(x-0)}dx\] \[=\left[ \frac{{{(x+1)}^{2}}}{2} \right]_{-1}^{0}+\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}=\frac{1}{2}+\frac{1}{2}=1\]You need to login to perform this action.
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