What is the area of the region bounded by the curve y=x^3-3x^2+2 and the x-axis?

Answer 1

What the question is asking is finding the area of the region bounded by the curves
f(x) = x^3 - 3x^2 + 2 and
g(x) = 0 (which is the x-axis).

To find the area, you must first determine their points of intersection. This is done by equating them to each other.

f(x) = g(x)

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

You'll find that this factors as

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

Equating each factor to 0,
x - 1 = 0 (which means x = 1)
x^2 - 2x - 2 = 0 (which means we'd have to use the quadratic formula to find the values).

x^2 - 2x - 2 = 0 means
x = [2 +/- sqrt(4 - 4(-2))]/2
x = [2 +/- sqrt(4 + 8)]/2
x = [2 +/- sqrt(12)]/2
x = [2 +/- 2sqrt(3)]/2
x = 1 +/- sqrt(3)

Our points of intersection are:
x = {1, 1 + sqrt(3), 1 - sqrt(3)}

Given that we have three points of intersection, we have to determine which is the higher curve between those points, and we will have *two* areas and two integrals to deal with.

Writing the intersecting points in increasing order, we have

x = {1 - sqrt(3), 1, 1 + sqrt(3)}

Test a point
a) between [1 - sqrt(3)] and 1,
b) between 1 and [1 + sqrt(3)]

a) Test x = 0; then f(0) = 0^3 - 3(0)^2 + 2 = 2, g(0) = 0,
so clearly f(x) is greater than g(x) in this interval.
b) Test x = 2; then f(2) = 2^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2.
g(2) = 0, so g(x) is higher in this interval.

The area is calculated by "higher curve" minus "lower curve".

Therefore,

A = Integral ([1 - sqrt(3)] to 1, [f(x) - g(x)] dx) +
Integral (1 to [1 + sqrt(3)], [g(x) - f(x)] dx )

Plugging in f(x) and g(x), we get

A = Integral ([1 - sqrt(3)] to 1, [x^3 - 3x^2 + 2 - 0] dx) +
Integral (1 to [1 + sqrt(3)], [0 - (x^3 - 3x^2 + 2)] dx)

A = Integral ([1 - sqrt(3)] to 1, [x^3 - 3x^2 + 2] dx) +
Integral (1 to [1 + sqrt(3)], (-x^3 + 3x^2 - 2) dx)

Integrate normally using the reverse power rule.

A = [(1/4)x^4 - x^3 + 2x] {evaluated from [1 - sqrt(3)] to 1} +
[(-1/4)(x^4) + x^3 - 2x] {evaluated from 1 to [1 + sqrt(3)] }

Plugging in our bounds for integration,

A = [ { (1/4)(1)^4 - 1^3 + 2(1) } - {(1/4)[1 - sqrt(3)]^4 -
[1 - sqrt(3)]^3 + 2[1 - sqrt(3)] } ] +
[ {(-1/4)[1 + sqrt(3)]^4 + [1 + sqrt(3)]^3 - 2[1 + sqrt(3)]} -
{(-1/4)(1)^4 + 1^3 - 2(1)} ]

Since this is getting algebraically ridiculous, I'm going to solve for certain values separately.

[1 - sqrt(3)]^4 = [ [1 - sqrt(3)]^2 ]^2 = [1 - 2sqrt(3) + 3]^2
= [4 - 2sqrt(3)]^2 = [16 - 16sqrt(3) + 12] = [28 - 16sqrt(3)]

[1 - sqrt(3)]^3 = [1 - sqrt(3)]^2 [1 - sqrt(3)] =
[1 - 2sqrt(3) + 3] ][1 - sqrt(3)] = [4 - 2sqrt(3)] [1 - sqrt(3)]
= [4 - 4sqrt(3) - 2sqrt(3) + 2(3)] = [4 - 6sqrt(3) + 6] =
[10 - 6sqrt(3)]

[1 + sqrt(3)]^4 = [ [1 + sqrt(3)]^2 ]^2 = [1 + 2sqrt(3) + 3]^2
= [4 + 2sqrt(3)]^2 = [16 + 16sqrt(3) + 12] = [28 + 16sqrt(3)]

[1 + sqrt(3)]^3 = [10 + 6sqrt(3)]

A = [ { (1/4) - 1 + 2 } - {(1/4) [28 - 16sqrt(3)] - [10 - 6sqrt(3)]
+ 2[1 - sqrt(3)]} ] +
[ {(-1/4)[28 - 16sqrt(3)] + [10 + 6sqrt(3)] - 2[1 + sqrt(3)]} - {(-1/4)(1)^4 + 1^3 - 2(1)} ]

And now it's a whole schwackload of simplification.

A = [ {5/4} - {7 - 4sqrt(3) - 10 + 6sqrt(3) + 2 - 2sqrt(3)} ] +
[ -7 + 4sqrt(3) + 10 + 6sqrt(3) - 2 - 2sqrt(3) + (1/4) - 1 + 2 ]

A = [ (5/4) - 7 + 4sqrt(3) + 10 - 6sqrt(3) - 2 + 2sqrt(3) ] +
[ 2 + 8sqrt(3) + (1/4) ]

A = [ (5/4) + 1 + 6sqrt(3)] + [2 + 8sqrt(3) + (1/4)]
A = 6/4 + 3 + 14sqrt(3)
A = 3/2 + 3 + 14sqrt(3)
A = 9/2 + 14sqrt(3)

{If there's an error it's an arithmetic one}

Answer 2

x^3-3x^2+2=0
x^3-3x^2+2=
(x-1)(x^2-2x-2)=
(x-1)[(x-1)^2-3]=
(x-1)*(x-1-sqrt(3))
*(x-1+sqrt(3))=0
thus
x=1
and
x=1+sqrt(3)
and
x=1-sqrt(3)
so
area=
abs(integral(x^3-3x^2+2)) in the interval [1-sqrt(3),1]
plus
abs(integral(x^3-3x^2+2)) in the interval [1,1+sqrt(3)]

and

integral(x^3-3x^2+2)=
(x^4)/4-x^3+2x

you should replace and find the answer,
sorry for my English.