What is the area bounded by y=x^2 and y=x^3-2x?
Euuw, sketch that
where do they hit?
x^2 = x^3 - 2x
x^3 -x^2 -2x = 0
x(x^2 -x -2) = 0
x (x-2)(x+1) = 0
so crashes at x = -1, 0 and 2
between -1 and 0, the cubic is on top
between 0 and 2, the parabola is on top
so do your integral from -1 to 0 of the cubic minus the parabola
then add the integral from 0 to 2 of the parabola minus the cubic.
so it should be 5/12+8/3=37/12?
must be done in 2 parts
total area
= ∫ (x^3 - 2x - x^2) dx from -1 to 0 + ∫(x^2 - x^3 + 2x) dx from 0 to 2
= [x^4/4 - x^2 - x^3/3]from -1 to 0 + [ x^3/3 - x^4/4 + x^2] from 0 to 2
= (0 - (1/4 - 1 - 1/3) + (8/3 - 4 + 4 - 0)
= 1 - 1/4 + 1/3 + 8/3
= 4 - 1/4
= 15/4
check my arithmetic
yes, 37/12 I get
(0 - (1/4 - 1 + 1/3) note plus 1/3
1 -1/4 -1/3
12/12 -3/12 - 4/12
5/12
then
5/12 +8/3
5/12 + 32/12
37/12
To find the area bounded by two curves, you need to find the points of intersection first. Then, you can integrate the difference between the two curves within those intersection points to calculate the area.
Let's find the points of intersection between the curves y = x^2 and y = x^3 - 2x:
x^2 = x^3 - 2x
Rearrange the equation:
x^3 - x^2 - 2x = 0
Factor out an x:
x(x^2 - x - 2) = 0
Set each factor equal to zero:
x = 0
x^2 - x - 2 = 0
To find the roots of the quadratic equation, you can either factor it or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic equation x^2 - x - 2 = 0, the values of a, b, and c are 1, -1, and -2, respectively. Substituting these values into the quadratic formula:
x = (-(-1) ± √((-1)^2 - 4(1)(-2))) / (2(1))
Simplifying:
x = (1 ± √(1 + 8)) / 2
x = (1 ± √9) / 2
x = (1 ± 3) / 2
So we have two possible values for x: x = -1 and x = 2.
Now that we have the intersection points, we can calculate the area by integrating the difference between the two curves within these boundaries.
The integral to find the area is given by:
Area = ∫[a, b] (f(x) - g(x)) dx
where f(x) is the upper curve and g(x) is the lower curve, and [a, b] are the intersection points.
In this case, f(x) = x^2 and g(x) = x^3 - 2x.
The area can be calculated as:
Area = ∫[-1, 2] ((x^2) - (x^3 - 2x)) dx
Simplifying:
Area = ∫[-1, 2] (x^2 - x^3 + 2x) dx
Evaluating this integral will give you the area bounded by the two curves y = x^2 and y = x^3 - 2x.