there is a line through the origin that divides the region bounded by the parabola $y = 7x - 3x^{2}$ and the…

Explanation:

Step1: Find the x - intercepts of the parabola

Set $y = 7x-3x^{2}=0$. Factor out an $x$: $x(7 - 3x)=0$. So the x - intercepts are $x = 0$ and $x=\frac{7}{3}$.

Step2: Calculate the area under the parabola

The area $A$ under the parabola $y = 7x-3x^{2}$ from $x = 0$ to $x=\frac{7}{3}$ is given by the definite integral $\int_{0}^{\frac{7}{3}}(7x - 3x^{2})dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: [ \begin{align*} \int_{0}^{\frac{7}{3}}(7x - 3x^{2})dx&=\left[\frac{7x^{2}}{2}-x^{3}\right]_{0}^{\frac{7}{3}}\ &=\frac{7}{2}\times\left(\frac{7}{3}\right)^{2}-\left(\frac{7}{3}\right)^{3}\ &=\frac{7^{3}}{2\times3^{2}}-\frac{7^{3}}{3^{3}}\ &=\frac{7^{3}}{3^{2}}\left(\frac{1}{2}-\frac{1}{3}\right)\ &=\frac{343}{9}\times\frac{1}{6}\ &=\frac{343}{54} \end{align*} ] So the area under the parabola is $\frac{343}{54}$, and the area of each of the two sub - regions is $\frac{343}{108}$.

Step3: Let the line be $y = mx$

Find the intersection point of $y = mx$ and $y = 7x-3x^{2}$. Set $mx=7x - 3x^{2}$. Rearrange to get $3x^{2}+(m - 7)x = 0$. Factor out an $x$: $x(3x+(m - 7))=0$. One solution is $x = 0$, and the other is $x=\frac{7 - m}{3}$.

Step4: Calculate the area between the line and the parabola

The area between the line $y = mx$ and the parabola $y = 7x-3x^{2}$ from $x = 0$ to $x=\frac{7 - m}{3}$ is given by the definite integral $\int_{0}^{\frac{7 - m}{3}}((7x - 3x^{2})-mx)dx$. Simplify the integrand: $(7x - 3x^{2})-mx=(7 - m)x-3x^{2}$. Integrate using the power - rule: [ \begin{align*} \int_{0}^{\frac{7 - m}{3}}((7 - m)x-3x^{2})dx&=\left[\frac{(7 - m)x^{2}}{2}-x^{3}\right]_{0}^{\frac{7 - m}{3}}\ &=\frac{(7 - m)}{2}\times\left(\frac{7 - m}{3}\right)^{2}-\left(\frac{7 - m}{3}\right)^{3}\ &=\left(\frac{7 - m}{3}\right)^{3}\left(\frac{1}{2}-\frac{1}{3}\right)\ &=\frac{(7 - m)^{3}}{54} \end{align*} ]

Step5: Solve for $m$

Since this area is equal to $\frac{343}{108}$, we set $\frac{(7 - m)^{3}}{54}=\frac{343}{108}$. Cross - multiply to get $2(7 - m)^{3}=343$. Then $(7 - m)^{3}=\frac{343}{2}$. Take the cube - root of both sides: $7 - m=\frac{7}{\sqrt[3]{2}}$. Solve for $m$: $m = 7-\frac{7}{\sqrt[3]{2}}=7\left(1-\frac{1}{\sqrt[3]{2}}\right)$.

Answer:

$7\left(1-\frac{1}{\sqrt[3]{2}}\right)$