Question

Consider the function f,g:
`\mathbb{R} \to\mathbb{R}` defined by


`\rm f(x) = {x}^(2) + (5)/(12) \: and \: g(x) = \begin{cases}2 \bigg( \rm 1 - (4 |x| )/(3) \bigg), \: \: \: \: |x| \leq (3)/(4), \\ \\ 0, \: \: \: \: \: \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: |x| > (3)/(4) .\end{cases}`
If
`\alpha` is the area of the region

`\rm \bigg \{(x,y) \in \mathbb{R} * \mathbb R : |x| \leq (3)/(4) ,0 \leq y \leq min \{f(x),g(x) \} \bigg \}`
then the value of 9
`\alpha`​

Answer 1

Suppose
`x>0`; then the curves meet when

`x^2 + \frac5{12} = 2 - \frac83 x \implies x^2 + \frac83 x - (19)/(12) = \left(x + \frac{19}6\right) \left(x - \frac12\right) = 0 \\\\ \implies x = \frac12`

By symmetry, they intersect at
`x=\pm\frac12`.

We also see that
`\min\left\\le\frac34\right\ = \frac5{12}` and
`\max\left\g(x): = 2` at
`x=0`.
`f` and
`g` are continuous, so it follows that

`\min\left\f(x),g(x) : = \begin{cases} f(x) & \text{if } |x| < \frac12 \\ g(x) & \text{if } |x| > \frac12 \\ f\left(\pm\frac12\right) = g\left(\pm\frac12\right) = \frac23 & \text{if } x = \pm\frac12 \end{cases}`

Compute the area
`\alpha`. Taking advantage of symmetry again, we have

`\alpha = \displaystyle 2 \int_0^(1/2) \int_0^(f(x)) dy\,dx + 2 \int_(1/2)^(3/4) \int_0^(g(x)) dy \, dx \\\\ ~~~~ = 2 \int_0^(1/2) f(x) \, dx + 2 \int_(1/2)^(3/4) g(x) \, dx \\\\ ~~~~ = \left. 2 \left(\frac{x^3}3 + (5x)/(12)\right) \right\vert_0^(1/2) + \left. 2 \left(2x - \frac{8x^2}6\right) \right\vert_(1/2)^(3/4) \\\\ ~~~~ = \left(\frac12 - 0\right) + \left(\frac32 - \frac43\right) = \frac23`

and it follows that
`9\alpha = \boxed{6}`.