A) 6 sq. unit
B) \[5/6\]sq. unit
C) \[1/6\]sq.unit
D) None of these
Correct Answer: C
Solution :
\[\frac{dy}{dx}=2x+1\Rightarrow y={{x}^{2}}+x+c\] |
Putting \[x=1,\,y=2,\] we get \[C=0\] \[\Rightarrow \,\,\,y={{x}^{2}}+x,\] |
\[\Rightarrow \,\,\,{{\left( x+\frac{1}{2} \right)}^{2}}=y+\frac{1}{4},\] which is a equation of parabola, whose vertices is \[\left( \frac{-1}{2},\frac{-1}{4} \right)\] |
\[\therefore \] required area \[=\left| \int\limits_{-1}^{0}{({{x}^{2}}+x)dx} \right|=\left( \frac{{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2} \right)_{-1}^{0}\]\[=\left| \frac{-1}{3}+\frac{1}{2} \right|=\frac{1}{6}\] sq. unit |
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