Answer to Question #334915 in Calculus for Quân Jason

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: $y=mx$

$6x-7x^2=mx$

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

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

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

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

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

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

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