If the line y=mx bisects the area enclosed by the lines x=0, y=0, x=32 and the curve y = 1…

Question

If the line $y=mx$ bisects the area enclosed by the lines $x=0$, $y=0$, $x=\dfrac32$ and the curve $y = 1+4x-x^2$, then $12m$ is equal to \_\_\_\_\_\_\_\_.

Answer: 26

Accepted Answer

$$\frac12\int_0^{3/2}(1+4x-x^2)\,dx = \int_0^{3/2}mx\,dx$$

$$\frac12\left[x+2x^2-\frac{x^3}{3}\right]_0^{3/2} = \frac{m}{2}\left[x^2\right]_0^{3/2}$$

$$\frac12\left(\frac32+\frac92-\frac98\right) = \frac{m}{2}\cdot\frac94$$

$$\frac32+\frac92-\frac98 = \frac{9m}{4} \implies \frac{39}{8}\cdot\frac{4}{9}=m \implies m=\frac{39}{18}=\frac{13}{6}$$

$12m = 26$

Solutions · 1