A curve y has gradient Dy/dx=3x^2-6x+2a)if the curve passes through the origin find it equation b)Find the area of the finite region included between the curve in a)and the x-axis.?
dy/dx=3x^2 - 6x + 2
y = x^3 - 3x^2 + 2x + c
but it passes throught (0,0)
0 = 0 - 0 + 0 + c -----> c = 0
y = x^3 - 3x^2 + 2x
b) y = x(x^2 - 3x + 2) = x(x - 1)(x - 2)
so the x intercepts are 0, 1, 2
giving you two enclosed areas between the curve and the x-axis
You will need ∫ x^3 - 3x^2 + 2x dx from 0 to 1 + ∫ -x^3 + 3x^2 - 2x dx from 1 to 2
You should have no difficulty with this straight-forward integral.
∫ x^3 - 3x^2 + 2x dx from 0 to 1 + ∫ -x^3 + 3x^2 - 2x dx from 1 to 2
To find the equation of the curve when it passes through the origin, we need to integrate the gradient function.
a) Integration of the gradient function:
∫(3x^2 - 6x + 2) dx
Using the power rule of integration:
= x^3 - 3x^2 + 2x + C
Since the curve passes through the origin (0,0), the constant term should be zero.
So, C = 0.
Therefore, the equation of the curve is:
y = x^3 - 3x^2 + 2x
b) To find the area of the finite region included between the curve and the x-axis, we need to find the integral of the absolute value of the curve between the points where it intersects the x-axis.
Since the curve passes through the origin (0,0), it intersects the x-axis at x = 0.
To find the other point where the curve intersects the x-axis, we set y = 0 in the equation of the curve and solve for x:
0 = x^3 - 3x^2 + 2x
Factoring out x:
0 = x(x^2 - 3x + 2)
Using the quadratic formula:
x^2 - 3x + 2 = 0
(x - 1)(x - 2) = 0
x = 1 or x = 2
So, the curve intersects the x-axis at x = 0, 1, and 2.
To find the area of the finite region, we need to integrate the absolute value of the curve:
∫|x^3 - 3x^2 + 2x| dx
We need to split the integral into two parts, one between x = 0 and x = 1, and the other between x = 1 and x = 2.
For x = 0 to x = 1:
∫(x^3 - 3x^2 + 2x) dx
Using the antiderivative:
= (1/4)x^4 - x^3 + x^2 | [0, 1]
= (1/4) - 1 + 1
= -3/4
For x = 1 to x = 2:
∫(x^3 - 3x^2 + 2x) dx
Using the antiderivative:
= (1/4)x^4 - x^3 + x^2 | [1, 2]
= (1/4)(2^4) - 2^3 + 2^2 - (1/4)(1^4) + 1^3 - 1^2
= 4 - 8 + 4 - (1/4) + 1 - 1
= 0
Therefore, the area of the finite region included between the curve and the x-axis is:
|-3/4 + 0| = 3/4
To find the equation of the curve, we need to integrate the given gradient function. Let's start with that.
a) Integrating Dy/dx = 3x^2 - 6x + 2:
∫(Dy/dx) dx = ∫(3x^2 - 6x + 2) dx
Integrating term by term:
∫Dy dx = ∫(3x^2 dx) - ∫(6x dx) + ∫(2 dx)
Integrating each term:
Y = x^3 - 3x^2 + 2x + C
Since the curve passes through the origin (0, 0), we can substitute the coordinates into the equation to find the value of C.
When x = 0, Y = 0:
0 = (0)^3 - 3(0)^2 + 2(0) + C
0 = C
Substituting C = 0 back into the equation, we get the equation of the curve:
Y = x^3 - 3x^2 + 2x
b) To find the area of the finite region included between the curve and the x-axis, we need to integrate the absolute value of the function.
The region is bounded by the x-axis and the curve, so we want to find the area between y = 0 and the curve y = x^3 - 3x^2 + 2x.
To calculate the area, integrate the absolute value of the curve equation:
Area = ∫|x^3 - 3x^2 + 2x| dx
This integral breaks into different segments where the curve intersects the x-axis.
As a first step, we need to find the x-values where y = 0:
0 = x^3 - 3x^2 + 2x
Factoring out an x:
0 = x(x^2 - 3x + 2)
Setting each factor equal to zero:
x = 0, x = 1, x = 2
We now have three intervals: [-∞, 0], [0, 1], [1, 2].
Using the absolute value function:
Area = ∫(0 to 1) (x^3 - 3x^2 + 2x) dx + ∫(1 to 2) -(x^3 - 3x^2 + 2x) dx
Integrating each segment and summing their absolute values gives the total area.
I hope this explanation helps you solve the problem. Let me know if you have any further questions!