y=x^3-4x^2+ 3x .Find the area between the curve and the X axis from X=0 and X=3
To find the area between the curve y=x^3-4x^2+3x and the x-axis from x=0 to x=3, we need to integrate the absolute value of the function from 0 to 3.
First, let's find the x-intercepts of the function:
y = x^3 - 4x^2 + 3x
0 = x^3 - 4x^2 + 3x
0 = x(x^2 - 4x + 3)
0 = x(x-1)(x-3)
So the x-intercepts are x=0, x=1, and x=3.
The function is negative between x=0 and x=1, and positive between x=1 and x=3. Therefore, we need to find the integral of the absolute value of the function between 0 and 3:
∫[0,3] |x^3-4x^2+3x| dx = ∫[0,1] -(x^3-4x^2+3x) dx + ∫[1,3] (x^3-4x^2+3x) dx
∫[0,1] -(x^3-4x^2+3x) dx = -∫[0,1] (x^3-4x^2+3x) dx = -[1/4 x^4 - 4/3 x^3 + 3/2 x^2] from 0 to 1
= -[1/4(1^4)-4/3(1^3)+3/2(1^2)] = -[1/4-4/3+3/2] = -[1/4 - 16/12 + 18/12] = -[3/12 - 16/12 + 18/12] = -[-5/12] = 5/12
∫[1,3] (x^3-4x^2+3x) dx = ∫[1,3] (x^3-4x^2+3x) dx = [1/4 x^4 - 4/3 x^3 + 3/2 x^2] from 1 to 3
= [1/4(3^4)-4/3(3^3)+3/2(3^2)] - [1/4(1^4)-4/3(1^3)+3/2(1^2)] = [81/4-108/3+27/2]- [1/4-4/3+3/2]
= [81/4-36-54/2] - [1/4-4/3+3/2] = [81/4-54-27]- [1/4 - 16/12 + 18/12] = [81/4-81]-[3/12] = 0-[3/12] = -3/12
Therefore, the total area between the curve and the x-axis from x=0 to x=3 is:
5/12 - 3/12 = 2/12 = 1/6 square units.