find the area of the region bounded by
y=4x, y=x^3, x=0 and x=2
Check this on paper, but in my mind I see the differential area (x^3-3x)dx
integrate that from 0 to 2
y=4x and y=x^3 intersect when x=-2,0 and 2
the vertical boundaries are x=0 and x=2, so
the area is the integral of 4x-x^3 from 0 to 2
which is [2x^2 - (1/4)x^4] from 0 to 2
=8-4 - 0
=4
To find the area of the region bounded by the given curves, you can follow these steps:
1. Determine the points of intersection between the curves.
- Set y=4x and y=x^3 equal to each other to find the x-values where the curves intersect:
4x = x^3
Rearrange the equation to get x^3 - 4x = 0
Factor out an x: x(x^2 - 4) = 0
Set each factor equal to zero: x = 0 and x^2 - 4 = 0
Solve for x: x = 0 and x = ±2
- The curves intersect at x = 0 and x = ±2.
2. Determine the horizontal and vertical boundaries.
- The horizontal boundary is given by the x-values where the curves intersect: x = 0 and x = 2.
- The vertical boundaries are y = 4x and y = x^3.
3. Set up the integral for the area.
- The area can be calculated by integrating the difference between the curves with respect to x, from the lower x-boundary to the upper x-boundary:
Area = ∫[lower boundary to upper boundary] (upper curve - lower curve) dx
In this case, the upper curve is y = 4x and the lower curve is y = x^3.
4. Evaluate the integral.
- The integral of (4x - x^3) with respect to x, from 0 to 2, can be calculated as follows:
∫[0 to 2] (4x - x^3) dx = [2x^2 - (1/4)x^4] evaluated from 0 to 2
Plugging in the upper and lower x-values:
= [2(2)^2 - (1/4)(2)^4] - [2(0)^2 - (1/4)(0)^4]
= (8 - 4) - (0 - 0)
= 4
Therefore, the area of the region bounded by y = 4x, y = x^3, x = 0, and x = 2 is 4 square units.