Let A denote the area enclosed by the equations $y = \sqrt [ 3 ] { x }$ and $y = 4 x$ Then A is

A) $\frac { 1 } { 64 }$
B) $\frac { 1 } { 32 }$
C) $\frac { 1 } { 16 }$
D) $\frac { 1 } { 8 }$
E) $\frac { 1 } { 4 }$

Question

Let A denote the area enclosed by the equations $y = \sqrt [ 3 ] { x }$ and $y = 4 x$ Then A is

A) $\frac { 1 } { 64 }$
B) $\frac { 1 } { 32 }$
C) $\frac { 1 } { 16 }$
D) $\frac { 1 } { 8 }$
E) $\frac { 1 } { 4 }$

Quizplus · Accepted Answer

The area enclosed by the equations is the area between the curves from the point of intersection to the right. To find the point of intersection, set the two equations equal to each other and solve for x:$\sqrt [ 3 ] { x } = 4 x$$\Rightarrow x^3 = 64 x$$\Rightarrow x^3 - 64 x = 0$$\Rightarrow x (x^2 - 64) = 0$$\Rightarrow x = 0$ or $x = 8$So the area enclosed by the equations is the area between the curves from $x = 0$ to $x = 8$.The area between the curves is given by the integral of the upper curve minus the integral of the lower curve:$A = \int_0^8 (4x) - (\sqrt [ 3 ] { x }) dx$$= \int_0^8 4x - x^{1/3} dx$$= \left [ 2x^2 - \frac { 3 } { 4 } x^{4/3} \right ]_0^8$$= 2(8)^2 - \frac { 3 } { 4 } (8)^{4/3}$$= 128 - 24$$= 104$So the area enclosed by the equations is 104 square units.

Let a Denote the Area Enclosed by the Equations y=x3y = \sqrt [ 3 ] { x }y=3x​

Let a Denote the Area Enclosed by the Equations $y = \sqrt [ 3 ] { x }$