A) \[\frac{5}{2}\]sq. unit
B) \[\frac{3}{2}\]sq. unit
C) \[\frac{1}{2}\]sq. unit
D) None of these
Correct Answer: A
Solution :
Required area \[\int_{-1}^{2}{\,y\,dx}=\int_{-1}^{0}{\,y\,.\,dx+\int_{0}^{2}{y\,.\,dx=\frac{5}{2}}}\]sq. unit.You need to login to perform this action.
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