If the line $$y = mx$$ bisects the area enclosed by the lines $$x = 0$$, $$y = 0$$, $$x = \frac{3}{2}$$ and the curve $$y = 1 + 4x - x^2$$, then $$12m$$ is equal to _________.

We are given the bounded region lying in the first quadrant under the curve $$y = 1 + 4x - x^{2}$$ and above the x-axis between the two vertical lines $$x = 0$$ and $$x = \dfrac{3}{2}$$. Its total area will be halved by some straight line through the origin $$y = mx$$. The requirement is that this line should divide the region into two parts of equal area.

The first task is to calculate the complete area enclosed by the four given boundaries. Because every ordinate in the strip from $$x = 0$$ to $$x = \dfrac{3}{2}$$ lies between $$y = 0$$ and $$y = 1 + 4x - x^{2}$$, the total area $$A$$ is obtained by the definite integral

$$ A \;=\; \int_{0}^{\frac{3}{2}} \bigl(1 + 4x - x^{2}\bigr)\,dx . $$

We integrate each term separately, using the standard formulas $$\displaystyle \int x^{n}\,dx = \frac{x^{n+1}}{n+1}\; (n \neq -1)$$ and $$\displaystyle \int k\,dx = kx$$.

$$ \begin{aligned} A &= \int_{0}^{\frac{3}{2}} 1\,dx \;+\; \int_{0}^{\frac{3}{2}} 4x\,dx \;-\; \int_{0}^{\frac{3}{2}} x^{2}\,dx \\[4pt] &= \Bigl[x\Bigr]_{0}^{\frac{3}{2}} + 4\Bigl[\frac{x^{2}}{2}\Bigr]_{0}^{\frac{3}{2}} - \Bigl[\frac{x^{3}}{3}\Bigr]_{0}^{\frac{3}{2}} \\[4pt] &= \left(\frac{3}{2} - 0\right) + 4\!\left(\frac{\bigl(\frac{3}{2}\bigr)^{2}}{2} - 0\right) - \left(\frac{\bigl(\frac{3}{2}\bigr)^{3}}{3} - 0\right) \\[4pt] &= \frac{3}{2} + 4\!\left(\frac{\frac{9}{4}}{2}\right) - \frac{\frac{27}{8}}{3} \\[4pt] &= \frac{3}{2} + 4\!\left(\frac{9}{8}\right) - \frac{27}{24} \\[4pt] &= \frac{3}{2} + \frac{9}{2} - \frac{9}{8} \\[4pt] &= \frac{12}{8} + \frac{36}{8} - \frac{9}{8} \\[4pt] &= \frac{39}{8}. \end{aligned} $$

So the region’s total area is $$A = \dfrac{39}{8}$$.

Next we consider the line $$y = mx$$. At any abscissa $$x$$ (with $$0 \le x \le \dfrac{3}{2}$$) the height of this line is $$mx$$, while the height of the curve is $$1 + 4x - x^{2}$$. We shall soon verify that the line always lies strictly below the curve throughout the required interval for the slope we finally obtain; hence the two figures that the line partitions are:

• the area under the line $$y = mx$$ from $$x=0$$ to $$x=\dfrac{3}{2}$$, and
• the remaining part of the region, which is the area between the curve and the line over the same $$x$$-range.

If the line bisects the region, the area under the line must therefore be exactly half the total area. We compute that area $$A_{\text{line}}$$ by integrating $$y = mx$$ from $$0$$ to $$\dfrac{3}{2}$$:

$$ A_{\text{line}} = \int_{0}^{\frac{3}{2}} mx\,dx = m \int_{0}^{\frac{3}{2}} x\,dx = m\left[\frac{x^{2}}{2}\right]_{0}^{\frac{3}{2}} = m\left(\frac{\bigl(\frac{3}{2}\bigr)^{2}}{2}\right) = m\left(\frac{\frac{9}{4}}{2}\right) = m\left(\frac{9}{8}\right) = \frac{9m}{8}. $$

The bisecting condition is now written as

$$ A_{\text{line}} \;=\; \frac{A}{2}. $$

Substituting the values just obtained,

$$ \frac{9m}{8} \;=\; \frac{1}{2}\!\left(\frac{39}{8}\right) \quad\Longrightarrow\quad \frac{9m}{8} \;=\; \frac{39}{16}. $$

We clear the denominators by multiplying both sides by $$16$$:

$$ 16 \times \frac{9m}{8} \;=\; 16 \times \frac{39}{16} \;\;\Longrightarrow\;\; 2 \times 9m \;=\; 39 \;\;\Longrightarrow\;\; 18m \;=\; 39. $$

Dividing by $$18$$ gives the slope:

$$ m \;=\; \frac{39}{18} \;=\; \frac{13}{6}. $$

Finally, the quantity asked for in the problem is $$12m$$. Multiplying, we find

$$ 12m \;=\; 12 \times \frac{13}{6} \;=\; 2 \times 13 \;=\; 26. $$

Thus the required number is $$26$$. So, the answer is $$26$$.