A) \[\frac{59}{6}\]
B) \[\frac{57}{4}\]
C) \[\frac{59}{3}\]
D) \[\frac{57}{6}\]
Correct Answer: A
Solution :
\[y={{x}^{2}}+3x=4\]\[\Rightarrow \] \[{{x}^{2}}+3x-4=0\] |
\[(x+4)(x-1)=0\] |
\[x=1\] |
area \[=\int\limits_{0}^{1}{({{x}^{2}}+3x)dx+2(4)}\]\[=\frac{1}{3}+\frac{3}{2}+8=\frac{11}{6}+8=\frac{59}{6}.\] |
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