Find the area (finite) enclosed by the two curves

Question

Find the area (finite) enclosed by the two curves Y=x^3 - 3x^2 + 3x and y= x. Sketch the region. Give both an exact value and decimal approx. to the nearest hundredth.

Answer 1

You want to start out by setting the equations equal to one another and finding out where they intersect. This will give you the lower and upper values of integration. You then want to integrate both functions and subtract the lower curve from the upper curve.

x^3-3x^2+3x=x

x^3-3x^2+2x=0

x(x^2-3x+2)=0

x(x-2)(x-1)=0

The curves intersect at x=0, x=1, and x=2.

From 0 to 1, the cubic function is positive while the linear function is negative with respect to finding area enclosed.

From 1 to 2, the linear function is positive while the cubic function is negative with respect to finding area enclosed.

In short, we want to integrate the functions from 0 to 1, then subtract the linear function from the cubic function. We then want to add to it the area created by integrating each function from 1 to 2, then subtract the cubic function from the linear function.

integral from (0,1) of x^3 - 3x^2 + 3x = (x^4)/4-x^3+(3x^2)/2 [0,1] = 1/4-1+3/2 = 1/4-4/4+6/4 = 3/4

Integral from (0,1) of x = (x^2)/2 [0,1] = 1/2-0=1/2

Therefore the area enclosed from 0 to 1 is 3/4-1/2 =1/4

integral from (1,2) of x^3 - 3x^2 + 3x = (x^4)/4-x^3+(3x^2)/2 [1,2] = (4-8+6)-(3/4)=1 1/4=5/4

Integral from (1,2) of x = (x^2)/2 [1,2] = 2-1/2 = (4/2)-(1/2)=3/2=6/4

Therefore the area enclosed from 1 to 2 is 6/4-5/4=1/4

Therefore the total area enclosed by the two curves is 1/4+1/4=2/4=1/2