Consider the functions $$f, g : \mathbb{R} \to \mathbb{R}$$ defined by $$f(x) = x^2 + \dfrac{5}{12}$$ and $$g(x) = \begin{cases} 2\left(1 - \dfrac{4|x|}{3}\right), & |x| \leq \dfrac{3}{4} \\ 0, & |x| > \dfrac{3}{4} \end{cases}$$

If $$\alpha$$ is the area of the region $$\left\{(x,y) \in \mathbb{R} \times \mathbb{R} : |x| \leq \dfrac{3}{4}, \, 0 \leq y \leq \min\{f(x), g(x)\}\right\}$$, then the value of $$9\alpha$$ is _______.

The required area is obtained by integrating the lower of the two curves $$f(x)$$ and $$g(x)$$ over the interval $$|x|\le\dfrac34$$:

$$\alpha=\int_{-3/4}^{3/4}\min\{f(x),g(x)\}\,dx.$$

First locate the points where the two graphs intersect inside $$[-\dfrac34,\dfrac34]$$.
For $$|x|\le\dfrac34$$ we have $$g(x)=2\!\left(1-\dfrac{4|x|}{3}\right).$$
Because both $$f(x)=x^{2}+\dfrac{5}{12}$$ and $$g(x)$$ are even, solve $$f(x)=g(x)$$ for $$x\ge0$$:

$$x^{2}+\dfrac{5}{12}=2-\dfrac{8x}{3}$$

$$\Longrightarrow\;x^{2}+\dfrac{8x}{3}-\dfrac{19}{12}=0.$$

Multiplying by $$12$$ gives $$12x^{2}+32x-19=0$$.
Using the quadratic formula:

$$x=\dfrac{-32\pm\sqrt{32^{2}-4\cdot12\cdot(-19)}}{24} =\dfrac{-32\pm\sqrt{1024+912}}{24} =\dfrac{-32\pm44}{24}.$$

The positive root is $$x=\dfrac12$$, which lies in $$\left[0,\dfrac34\right]$$.
Hence the curves meet at $$x=\pm\dfrac12$$.

Check which curve is lower between these points.
At $$x=0$$: $$f(0)=\dfrac{5}{12}\lt g(0)=2\;,$$ so $$f(x)$$ lies below $$g(x)$$ near the origin.
At $$x=\dfrac34$$: $$f\!\left(\dfrac34\right)=\dfrac{9}{16}+\dfrac{5}{12}\approx0.984\;, \;g\!\left(\dfrac34\right)=0,$$ so $$g(x)$$ is lower near the ends.
Therefore

• For $$|x|\le\dfrac12$$, the lower curve is $$f(x)$$.
• For $$\dfrac12\le|x|\le\dfrac34$$, the lower curve is $$g(x)$$.

Because of evenness, integrate over $$[0,\dfrac34]$$ and double:

$$\alpha=2\left[\int_{0}^{1/2}f(x)\,dx+\int_{1/2}^{3/4}g(x)\,dx\right].$$

First integral:
$$\int_{0}^{1/2}\left(x^{2}+\dfrac{5}{12}\right)dx =\left[\dfrac{x^{3}}{3}\right]_{0}^{1/2}+\left[\dfrac{5x}{12}\right]_{0}^{1/2} =\dfrac{1}{24}+\dfrac{5}{24}=\dfrac14.$$

Second integral: for $$1/2\le x\le3/4$$, $$g(x)=2-\dfrac{8x}{3}$$.

$$\int_{1/2}^{3/4}\left(2-\dfrac{8x}{3}\right)dx =\left[2x-\dfrac{4x^{2}}{3}\right]_{1/2}^{3/4}.$$

At $$x=\dfrac34$$: $$2\!\left(\dfrac34\right)-\dfrac{4}{3}\!\left(\dfrac{3}{4}\right)^{2} =\dfrac32-\dfrac34=\dfrac34.$$
At $$x=\dfrac12$$: $$2\!\left(\dfrac12\right)-\dfrac{4}{3}\!\left(\dfrac12\right)^{2} =1-\dfrac13=\dfrac23.$$

Difference: $$\dfrac34-\dfrac23=\dfrac{9-8}{12}=\dfrac1{12}.$$

Total area:

$$\alpha=2\left[\dfrac14+\dfrac1{12}\right] =2\left[\dfrac3{12}+\dfrac1{12}\right] =2\!\left(\dfrac13\right)=\dfrac23.$$

Finally, $$9\alpha=9\left(\dfrac23\right)=6.$$

Hence the required value is 6.