Question

There is a line through the origin that divides the region bounded by the parabola y=2x-4x^2 and the x-axis into two regions with equal area. What is the slope of that line? ...?

Answer 1

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y = 7x - 4x² 

7x - 4x² = 0 

x(7 - 4x) = 0 

x = 0, 7/4 

Find the area of the bounded region... 

A = ∫ 7x - 4x² dx |(0 to 7/4) 

A = 7/2 x² - 4/3 x³ |(0 to 7/4) 

A = 7/2(7/4)² - 4/3(7/4)³ - 0 = 3.573 

Half of this area is 1.786, now set up an integral that is equal to this area but bounded by the parabola and the line going through the origin... 

y = mx + c 

c = 0 since it goes through the origin 

The point where the line intersects the parabola we shall call (a, b) 

y = mx ===> b = m(a) 

Slope = m = b/a 

Now we need to integrate from 0 to a to find the area bounded by the parabola and the line... 

1.786 = ∫ 7x - 4x² - (b/a)x dx |(0 to a) 

1.786 = (7/2)x² - (4/3)x³ - (b/2a)x² |(0 to a) 

1.786 = (7/2)a² - (4/3)a³ - (b/2a)a² - 0 

1.786 = (7/2)a² - (4/3)a³ - b(a/2) 

Remember that (a, b) is also a point on the parabola so y = 7x - 4x² ==> b = 7a - 4a² 
Substitute... 

1.786 = (7/2)a² - (4/3)a³ - (7a - 4a²)(a/2) 

1.786 = (7/2)a² - (4/3)a³ - (7/2)a² + 2a³ 

(2/3)a³ = 1.786 

a = ∛[(3/2)(1.786)] 

a = 1.39 

b = 7(1.39) - 4(1.39)² = 2.00 

Slope = m = b/a = 2.00 / 1.39 = 1.44