find the area of the region bounded by the curves f(x)=x-x^3 ; g(x)=x^2-x ; over [0,1]
since f > g on (0,1), the area is just
∫[0,1] (x-x^3)-(x^2-x) dx = 5/12
ty!
i got 5/6?
i used the equation 2x-x^5, is that wrong?
it sure is
(x-x^3)-(x^2-x) = -x^3-x^2+2x
You can't combine x^2 and x^3 to make x^5!! They're added, not multiplied.
To find the area of the region bounded by the curves f(x) and g(x) over the interval [0, 1], you first need to determine the points of intersection between the two curves. These points will define the boundaries of the region.
For this problem, let's find the points where f(x) = g(x):
x - x^3 = x^2 - x
To simplify, let's move all the terms to one side:
x^3 - x^2 + 2x - x = 0
Simplifying further:
x^3 - x^2 + x = 0
Now we can factor out an x:
x(x^2 - x + 1) = 0
The quadratic equation x^2 - x + 1 does not have real solutions, so the only point of intersection between the two curves is when x = 0.
To calculate the area of the region, we need to integrate the difference between the functions f(x) and g(x) within the interval [0, 1].
Let's set up the integral:
Area = ∫[0,1] (f(x) - g(x)) dx
Substituting the functions:
Area = ∫[0,1] (x - x^3 - (x^2 - x)) dx
Simplifying further:
Area = ∫[0,1] (x - x^2 + x^3 - x) dx
Now, integrate the expression:
Area = ∫[0,1] (x^3 - x^2) dx
Integrating:
Area = (1/4)x^4 - (1/3)x^3 | [0,1]
Substituting the limits of integration:
Area = (1/4)(1^4 - 0^4) - (1/3)(1^3 - 0^3)
Area = (1/4) - (1/3)
Area = 1/12 square units
Therefore, the area of the region bounded by the curves f(x) = x - x^3 and g(x) = x^2 - x over the interval [0, 1] is 1/12 square units.