A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
\[\int_{-2}^{2}{|[x]|}\,dx=\int_{\,-2}^{\,-1}{\,|[x]|dx+\int_{-1}^{0}{|[x]|dx+\int_{0}^{1}{|[x]|dx|+\int_{1}^{2}{|[x]|dx}}}}\] \[=\int_{-2}^{-1}{2dx\,\,}+\int_{-1}^{0}{1dx+\int_{0}^{1}{0\,dx+}}\int_{1}^{2}{1dx}\] \[=2[x]_{-2}^{-1}+[x]_{-1}^{0}+0+[x]_{1}^{2}\] \[=2(-1+2)+(0+1)+(2-1)=2+1+1=4.\]You need to login to perform this action.
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