A) 4
B) 3
C) 2
D) 1
Correct Answer: C
Solution :
[c] :\[I=\int\limits_{-2}^{2}{(x-[x])dx}=\int\limits_{-2}^{-1}{(x+2)dx}+\int\limits_{-1}^{0}{(x+1)}dx\] \[+\int\limits_{0}^{1}{x\,dx}+\int\limits_{1}^{2}{(x-1)\,}dx\] \[=\left[ \frac{{{x}^{2}}}{2}+2x \right]_{-2}^{-1}+\left[ \frac{{{x}^{2}}}{2}+x \right]_{-1}^{0}+\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}+\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}\] \[=\frac{1}{2}-2-(2-4)+0-\left( \frac{1}{2}-1 \right)+\frac{1}{2}+2-2-\left( \frac{1}{2}-1 \right)\]\[=2\]
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