A) 0 sq unit
B) 3/2 sq unit
C) 5/2 sq units
D) - 5/2 sq unit
Correct Answer: C
Solution :
Area of the region bounded by the curve\[y=x,\] \[x-\]axis and abscissae\[x=-1\]and\[x=2\]is given by \[A=\int_{-1}^{2}{x\,dx}=\left| \int_{-1}^{0}{x\,dx} \right|+\left| \int_{0}^{2}{x\,dx} \right|\] \[=\left| \left[ \frac{{{x}^{2}}}{2} \right]_{-1}^{0} \right|+\left| \left[ \frac{{{x}^{2}}}{2} \right]_{0}^{2} \right|\] \[=\left| 0-\frac{1}{2} \right|+\left| \frac{4}{2}-0 \right|\] \[=\frac{1}{2}+2\] \[=\frac{5}{2}\]sq units.
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