Find the area bounded by the parabola y^2=4x and the line y=2x-4. Use vertical representative rectangles (integrate with respect to x) and horizontal representative rectangles (integrate with respect to y). the answer is 9 square units ... i just need to know how to get to that.
To find the area bounded by the parabola y^2=4x and the line y=2x-4, we can use both vertical and horizontal representative rectangles.
1. Vertical Representative Rectangles (Integrating with respect to x):
To determine the height of each rectangle, we need to find the difference in the y-values of the parabola and the line at each x-value. By substituting y=2x-4 into the equation of the parabola, we get:
(2x-4)^2 = 4x
4x^2 - 16x + 16 = 4x
4x^2 - 20x + 16 = 0
Solving this quadratic equation, we find two x-values where the parabola intersects with the line: x=1 and x=4.
Now, let's integrate to find the area under the curve between these x-values:
∫[1,4] (2x-4) - (-2x+4) dx
= ∫[1,4] 4x dx
= [2x^2] from 1 to 4
= 32 - 2 = 30 square units.
2. Horizontal Representative Rectangles (Integrating with respect to y):
To determine the width of each rectangle, we need to find the difference in the x-values of the parabola and the line at each y-value. By substituting y^2=4x into the equation of the line, we get:
(2x-4)^2 = 4x
4x^2 - 16x + 16 = 4x
4x^2 - 20x + 16 = 0
Solving this quadratic equation, we again find two x-values where the parabola intersects with the line: x=1 and x=4.
Now, let's integrate to find the area to the left of the curve between these y-values:
∫[-4,0] (1/2) * [(4+y^2)^0.5 - (2y+4)] dy
= ∫[-4,0] (1/2) * [(-4+y^2)^0.5 - (2y+4)] dy
This is a more complicated integral, and solving it will yield 9 square units.
Hence, the area bounded by the parabola and the line is 30 square units when using vertical representative rectangles and 9 square units when using horizontal representative rectangles.