Calculus II, finding the area between two lines

Question

There is a line through the origin that divides the region bounded by the parabola y=4x-7x² and the x-axis into two regions with equal area. What is the slope of that line?

Answer 1

Find the area of the entire region:

A = ∫₀^4/7 (4x - 7x²)dx

= -7/3x³ + 2x² ] ₀^4/7 = - 64/147 + 32/49 = 32/147

1/2A = 16/147

The line y = mx intersects the given parabola when mx = 4x - 7x².

7x² - 4x + mx = 0 ; x = 0 or x = (4 - m)/7

∫₀^(4-m)/7 [-7x² + 4x - mx] dx

= -7/3x³ + 2x² - m/2x² ]₀^(4-m)/7

= -(4-m)³/147 + 2(4-m)²/49 - m(4-m)²/98

= [-3m/2(4-m)² + 6(4-m)² - (4-m)³] / 147 = 16 /147

(4-m)² [ -3m/2 + 6 - 4 + m] = 16

(4-m)² [ - m/2 + 2] = 16

[m² - 8m + 16][ - m/2 + 2] = 16

-1/2m³ + 6m² - 24m +16 = 0

m³ - 12m² + 48m - 32 = 0

m ~ .825