There is a line through the origin that divides the region bounded by the parabola y=3x-5x^2 and the x-axis into two regions with equal area. What is the slope of that line?

Hi James,y = 3x - 5x2 = x(3 - 5x) crosses the x-axis at x = 0 and x = 5/3 so find the area of the region bounded by y = 3x - 5x2 and the x-axis by integrating 3x - 5x2 from x = 0 to x = 5/3. Call the value A.Suppose the line through the prigin y = mx intersects the parabola at Q with x-coordinate p.

The task is to find m so that

Evaluation of the integral gave me1/2 (3 -m) p2 - 5/3 p3 = A/2 (1)The relationship between p and m comes from the fact that the point Q is on the line y = mx and also on the parabola y = 3x - 5x2. Thus mp = 3p - 5p2 or p = (3 - m)/5. Substitute into (1) to get1/2 (3 - m) (3 - m)2/25 -5/3 (3 - m)3/125 = A/2Solve for m.Harley

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It will help if you draw a rough graph.

The parabola is y = -2x(x-4), so it crosses the x-axis at (0,0) and again at (4,0). So to first find the total area under the parabola, integrate 8x - 2x^2 dx from x=0 to 4.

A line that passes through the origin has an equation of y=mx. We just need to find the value of m such that the line divides the original area into two pieces of equal area.

The line intersects the parabola when mx = 8x - 2x^2 2x^2 + mx - 8x = 0 2x^2 + (m-8)x = 0 x (2x + (m-8)) = 0 So this happens when x=0 (which we already know; both the parabola and the line pass through the origin), and when 2x+m-8=0, which is when x=(8-m)/2.

Consider the area bounded on the top left by the line y=mx, on the top right by the parabola from x=(8-m)/2 to x=4, and on the bottom by the x-axis. If you draw a vertical line through x=(8-m)/2, this splits the area up into a triangular piece on the left and a curved piece on the right. The area of this is mx dx + 8x-2x^2 dx where the first integral is taken from x=0 to x=(8-m)/2, an the second integral is taken from x=(8-m)/2 to x=4.

Calculate this. Now set it equal to HALF of the total area you found earlier. Solve for m.

1) First let us find the zeros of the given parabola function; so, 5x - 3x^2 = 0 Solving this, x = 0 & x = 5/3

2) The area of this region (that is enclosed by the curve, x-axis and the ordinates x = 0 & x = 5/3) is given by A = (5x - 3x^2) (dx) in limits (from 0 to 5/3) sq units

==> A = (5x^2/2 - x^3) in (0 to 5/3); evaluating for the upper limit 5/3 and lower limit 0,

A = 125/54 sq units. ----- (1)

3) Let the line passing through the origin is y = mx. [Since passing through the origin, the y intercept is zero].

4) Solving the line y = mx and the parabola, y = 5x - 3x^2, we have, 3x^2 - 5x + mx = 0 ==> x = 0 or x = (5-m)/3

4) Thus the area A obtained in equation (1) above, is divided into two equal parts by the line y = mx, within the limits x = 0 to x = (5-m)/3

5) Hence, area of the region A2 = (5x - 3x^2) - (mx) dx in 0 to (5-m)/3, = (1/2)(A)

==> {(5x^2)/2 - (mx^2)/2 - x^3} in 0 to (5-m)/3 = (1/2)(125/54)

On evaluating for the given limits and simplifying we get,

{(5-m)^3}/54 = 125/108

==> (5-m)^3 = 125/2

==> Thus m = 5 - {5/2^(1/3)}

[Solving the above, the slope m is nearly 1.032}

1. The parabola will intersect the x-axis at x = 0 and x = 7/8.

2. Integrate 7x - 8x^2 from 0 to 7/8. That's your "whole area". 0.893

3. Your line has equation y = mx. [you want m].

--- Solve mx = 7x - x^2 for x. Your two solutions are x = 0 and x = (7-m)/8

4. Integrate (7x - 8x^2) - (mx) from 0 to (7-m)/8.

--- You get an answer that contains m.

5. Set that answer equal to "whole area" / 2 and solve for m.

--- I get something like m = 1.444

--- Your intersection is x = (7 - m)/8 = 0.6945

That should do it.