question_answer

The area bounded by the curve\[y\le {{x}^{2}}+3x,0\le 4,\]\[0\le x\le 3,\]is:

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 \[=\int\limits_{0}^{1}{({{x}^{2}}+3x)dx+2(4)}\]\[=\frac{1}{3}+\frac{3}{2}+8=\frac{11}{6}+8=\frac{59}{6}.\]