Definite Integration Question 435
Question: Area lying in the first quadrant and bounded by the curve $ y=x^{3} $ and the line y = 4x is
Options:
A) 2
B) 3
C) 4
D) 5
Show Answer
Answer:
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})={{( 2x^{2}-\frac{x^{4}}{4} )}_0}^{2}=4} $
BETA
