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