question_answer

Area lying in the first quadrant and bounded by the curve \[y={{x}^{3}}\] and the line y = 4x is

A) 2                     

B) 3

C) 4         

D) 5

Correct Answer: C

Solution :

[c] The fine \[y=4x\]meets \[y={{x}^{3}}\]at \[4x={{x}^{3}}.\] \[\therefore x=0,\,\,2,\,-2\therefore y=0,8,-8\] \[\therefore \] Area in first quadrant \[=\int_{0}^{2}{(\underset{L:}{\mathop{y}}\,-\underset{C}{\mathop{{{y}_{2}}}}\,)}dx\] \[A=\int_{0}^{2}{(4x-{{x}^{3}})={{\left( 2{{x}^{2}}-\frac{{{x}^{4}}}{4} \right)}_{0}}^{2}=4}\]