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Mathematics
Mathematics
Problem
Consider the function
f
,
g
:
R
→
R
f, g: \mathbb{R} \to \mathbb{R}
f
,
g
:
R
→
R
defined by
f
(
x
)
=
x
2
+
5
12
f(x) = x^2 + \frac{5}{12}
f
(
x
)
=
x
2
+
12
5
and
g
(
x
)
=
{
2
(
1
−
4
∣
x
∣
3
)
,
∣
x
∣
≤
3
4
,
0
,
∣
x
∣
>
3
4
g(x) = \begin{cases} 2\left(1 - \frac{4|x|}{3}\right), & |x| \le \frac{3}{4}, \\ 0, & |x| > \frac{3}{4} \end{cases}
g
(
x
)
=
{
2
(
1
−
3
4∣
x
∣
)
,
0
,
∣
x
∣
≤
4
3
,
∣
x
∣
>
4
3
. If
α
\alpha
α
is the area of the region
{
(
x
,
y
)
∈
R
×
R
:
∣
x
∣
≤
3
4
,
0
≤
y
≤
min
{
f
(
x
)
,
g
(
x
)
}
}
\left\{(x, y) \in \mathbb{R} \times \mathbb{R} : |x| \le \frac{3}{4}, 0 \le y \le \min\{f(x), g(x)\}\right\}
{
(
x
,
y
)
∈
R
×
R
:
∣
x
∣
≤
4
3
,
0
≤
y
≤
min
{
f
(
x
)
,
g
(
x
)}
}
, then the value of
9
α
9\alpha
9
α
is \_\_\_\_\_\_.
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