Question

There is a line through the origin that divides the region bounded by the parabola 

y = 6x - 7x² and the x-axis into two regions with equal area. What is the slope of that line?

Answer 1

$$6x-7x^2=0$$

$$x_1=0, x_2=\dfrac{6}{7}$$

$$Area_{total}=\displaystyle\int_{0}^{6/7}(6x-7x^2)dx$$

$$=[3x^2-\dfrac{7}{3}x^3]\begin{matrix} 6/7 \\ 0 \end{matrix}=\dfrac{36}{49}({units}^2)$$

$$A_1=A_2=\dfrac{18}{49}{units}^2$$

Let $m=$ the slope of a line: $y=mx$

$$6x-7x^2=mx$$

$$x_1=0, x_2=\dfrac{6-m}{7}$$

$$Area_1=\displaystyle\int_{0}^{(6-m)/7}(6x-7x^2-mx)dx$$

$$=[3x^2-\dfrac{7}{3}x^3-\dfrac{m}{2}x^2]\begin{matrix} (6-m)/7 \\ 0 \end{matrix}$$

$$=\dfrac{3(6-m)^2}{49}-\dfrac{(6-m)^3}{3(49)}-\dfrac{m(6-m)^2}{2(49)}=\dfrac{18}{49}$$

$$(6-m)^2(18-12+2m-3m)=108$$

$$6-m=\sqrt[3]{108}$$

$$m=6-3\sqrt[3]{4}$$