How do you find the smaller area bounded by #y=4x-x^3# and #y=x^2-2x#?

Question

Emily Ammerman · Accepted Answer

#5.33" units"^2# ( 2 .d.p.)

From the graph we can see that the area we seek is in the interval #[0,2]#

For accuracy we need to find the points of intersection algebraically rather than trying to find them from the graph.

If:

#y=x^2-2x# and #y=4x-x^3#

Then:

#x^2-2x=4x-x^3=>x^3+x^2-6x=0#

Solving for #x#:

#x(x^2-6x)=0=>x(x+3)(x-2)=0#

Points of intersection:

#(0,0) , (2,0) , (-3,15)#

Area under #4x-x^3# in #[0,2]#

#"A"=int_(0)^(2)(4x-x^3) dx=2x^2-1/4x^4=[2x^2-1/4x^4]_(0)^(2)#

#[2x^2-1/4x^4]^(2)-[2x^2-1/4x^4]_(0)#

Plugging in upper and lower bounds:

#[2(2)^2-1/4(2)^4]^(2)-[2(0)^2-1/4(0)^4]_(0)#

#[8-4]^(2)-[0]_(0)#

Area #= 4#

Area under #x^2-2x# in #[0,2]#

#"A"=int_(0)^(2)(x^2-2x)dx=1/3x^3-x^2=[1/3x^3-x^2]_(0)^(2)#

#[1/3x^3-x^2]^(2)-[1/3x^3-x^2]_(0)#

Plugging in upper and lower bounds:

#[1/3(2)^3-(2)^2]^(2)-[1/3(0)^3-(0)^2]_(0)#

#[8/3-4]^(2)-[0]_(0)=-4/3#

Area #= 4/3#

Total area:

#4+4/3=16/3=5.33" units"^2# ( 2 .d.p)

Scarlett Allison · Answer

undefined To find the smaller area bounded by the two curves, you need to determine the points where they intersect. Set the equations equal to each other and solve for x to find the x-coordinates of the intersection points. Then, integrate the absolute difference between the two curves with respect to x over the interval where one curve is above the other. The smaller area will be the result of this integration.