It is given that $$f(x)=\begin{cases}4x,&0\le x<1\\(3-x)^2,&1\le x\le 3\end{cases}$$ and $f(-x)=-f(x)$. It is also known that $f(x)=f(x+6)$ for all real values of $x$.

1. Relate $f(4)$ to $f(-2)$ using the periodic property.

Given $f(x) = f(x+6)$ for all real $x$, we can substitute $x = -2$:
$f(4) = f(-2 + 6) = f(-2)$

2. Evaluate $f(-2)$ using the odd function property and the piecewise definition.

Given $f(-x) = -f(x)$, we have $f(-2) = -f(2)$.
For $x = 2$, we use the interval $1 \le x \le 3$ where $f(x) = (3-x)^2$.
$f(2) = (3 - 2)^2 = 1$
Therefore, $f(4) = -f(2) = -1$ (shown).

1. Sketch the primary piecewise function for $0 \le x \le 3$.

For $0 \le x < 1$, the graph of $y = 4x$ is a straight line segment starting from $(0,0)$ and ending at $(1,4)$.
For $1 \le x \le 3$, the graph of $y = (3-x)^2$ is a parabola opening upwards. It starts at $(1,4)$ and has its vertex at the $x$-axis intercept $(3,0)$.

2. Apply the odd function property to extend the graph to $-3 \le x < 0$.

The condition $f(-x) = -f(x)$ implies the graph has $180^\circ$ rotational symmetry about the origin.
Reflecting the primary graph through the origin gives:

• A straight line from $(0,0)$ down to $(-1,-4)$.
• A downward-opening parabola from $(-1,-4)$ ending at its vertex $(-3,0)$.
  This forms a complete period of length $6$ on the interval $[-3, 3]$.

3. Use the periodic property to extend the graph over the required full domain $-6 \le x \le 9$.

The relation $f(x) = f(x+6)$ indicates the function is periodic with a period of $6$.
Translate the basic shape of $[-3, 3]$ to the left by $6$ units for $[-6, -3]$:

• $(0,0) \rightarrow (-6,0)$
• $(1,4) \rightarrow (-5,4)$
  Translate the shape of $[-3, 3]$ to the right by $6$ units for $[3, 9]$:
• $(-1,-4) \rightarrow (5,-4)$
• $(0,0) \rightarrow (6,0)$
• $(1,4) \rightarrow (7,4)$
• $(3,0) \rightarrow (9,0)$
  Connect these translated key points with the corresponding alternating straight lines and parabolic curves.

1. Evaluate the integral of $|f(x)|$ over $0 \le x \le 3$.

Since $f(x) \ge 0$ for $0 \le x \le 3$, $|f(x)| = f(x)$ on this interval.
$\int_0^3 |f(x)| \, \mathrm{d}x = \int_0^1 4x \, \mathrm{d}x + \int_1^3 (3-x)^2 \, \mathrm{d}x$
$\int_0^3 |f(x)| \, \mathrm{d}x = \left[ 2x^2 \right]_0^1 + \left[ -\frac{(3-x)^3}{3} \right]_1^3$
$\int_0^3 |f(x)| \, \mathrm{d}x = (2 - 0) + \left( 0 - \left(-\frac{8}{3}\right) \right)$
$\int_0^3 |f(x)| \, \mathrm{d}x = 2 + \frac{8}{3} = \frac{14}{3}$

2. Use the odd property of $f(x)$ to evaluate the integral over a full period $[-3, 3]$.

Given $f(-x) = -f(x)$, we have $|f(-x)| = |-f(x)| = |f(x)|$, making $|f(x)|$ an even function.
The area under the graph is symmetric about the $y$-axis, so the integral over the half-period $[-3, 0]$ is equal to the integral over $[0, 3]$.
The integral over one full period of 6 (from $x = -3$ to $x = 3$) is:
$\int_{-3}^3 |f(x)| \, \mathrm{d}x = 2 \int_0^3 |f(x)| \, \mathrm{d}x = 2 \left(\frac{14}{3}\right) = \frac{28}{3}$

3. Apply the periodic property to evaluate the integral over $[-5, 7]$.

Since $f(x+6) = f(x)$, the function $|f(x)|$ is also periodic with a period of 6.
The interval from $x = -5$ to $x = 7$ has a length of $7 - (-5) = 12$, which comprises exactly two full periods.
For any periodic function, the integral over an interval spanning multiple full periods is independent of the starting point.
$\int_{-5}^7 |f(x)| \, \mathrm{d}x = 2 \int_{-3}^3 |f(x)| \, \mathrm{d}x$
$\int_{-5}^7 |f(x)| \, \mathrm{d}x = 2 \left(\frac{28}{3}\right) = \frac{56}{3}$