Find the total area between $$ f(x) $$ and the $$ x $$-axis from $$ x=0 $$ to $$ x=6 $$ $$ f(x)=x^{2}-x-12 $$ Total Area $$ =[ $$ ?] Round your answer to the nearest thousandth.

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Let $$ f(x) = x^2 - x - 12 $$. We want the total area between $$ f(x) $$ and the $$ x $$-axis from $$ x = 0 $$ to $$ x = 6 $$.

### 1. Find the Roots

Solve for $$ x $$ when
$$
x^2 - x - 12 = 0.
$$
Factor:
$$
(x - 4)(x + 3) = 0,
$$
so the roots are
$$
x = 4 \quad \text{and} \quad x = -3.
$$
Within the interval $$ [0, 6] $$, the only relevant root is $$ x = 4 $$.

### 2. Determine the Sign of $$ f(x) $$

Test a point in each sub-interval:
- For $$ x $$ in $$ [0, 4] $$, choose $$ x = 2 $$:
  $$
  f(2) = 2^2 - 2 - 12 = 4 - 2 - 12 = -10 \quad (\text{negative}).
  $$
- For $$ x $$ in $$ [4, 6] $$, choose $$ x = 5 $$:
  $$
  f(5) = 5^2 - 5 - 12 = 25 - 5 - 12 = 8 \quad (\text{positive}).
  $$

Thus, $$ f(x) $$ is below the $$ x $$-axis from $$ x = 0 $$ to $$ x = 4 $$ and above the $$ x $$-axis from $$ x = 4 $$ to $$ x = 6 $$.

### 3. Set Up the Integrals for the Total Area

The total area is computed by integrating the absolute value:
$$
\text{Total Area} = \int_{0}^{4} \lvert f(x) \rvert \, dx + \int_{4}^{6} \lvert f(x) \rvert \, dx.
$$
Since $$ f(x) $$ is negative on $$ [0,4] $$,
$$
\lvert f(x) \rvert = -f(x) \quad \text{on } [0,4],
$$
and positive on $$ [4,6] $$,
$$
\lvert f(x) \rvert = f(x) \quad \text{on } [4,6].
$$
Thus,
$$
\text{Total Area} = -\int_{0}^{4} (x^2 - x - 12) \, dx + \int_{4}^{6} (x^2 - x - 12) \, dx.
$$

### 4. Compute the Antiderivative

Find the antiderivative of $$ f(x) $$:
$$
\int (x^2 - x - 12) \, dx = \frac{x^3}{3} - \frac{x^2}{2} - 12x + C.
$$

### 5. Evaluate the Integrals

#### First Integral: from $$ x = 0 $$ to $$ x = 4 $$
$$
-\int_{0}^{4} (x^2 - x - 12) \, dx = -\left[ \left( \frac{x^3}{3} - \frac{x^2}{2} - 12x \right) \Bigg|_{0}^{4} \right].
$$
At $$ x = 4 $$:
$$
\frac{4^3}{3} - \frac{4^2}{2} - 12(4) = \frac{64}{3} - \frac{16}{2} - 48 = \frac{64}{3} - 8 - 48 = \frac{64}{3} - 56.
$$
At $$ x = 0 $$, the antiderivative is $$ 0 $$. Thus,
$$
-\left[\frac{64}{3} - 56\right] = 56 - \frac{64}{3} = \frac{168 - 64}{3} = \frac{104}{3}.
$$

#### Second Integral: from $$ x = 4 $$ to $$ x = 6 $$
$$
\int_{4}^{6} (x^2 - x - 12) \, dx = \left( \frac{x^3}{3} - \frac{x^2}{2} - 12x \right) \Bigg|_{4}^{6}.
$$
At $$ x = 6 $$:
$$
\frac{6^3}{3} - \frac{6^2}{2} - 12(6) = \frac{216}{3} - \frac{36}{2} - 72 = 72 - 18 - 72 = -18.
$$
At $$ x = 4 $$ (computed previously):
$$
\frac{4^3}{3} - \frac{4^2}{2} - 12(4) = \frac{64}{3} - 56.
$$
Thus, the second integral is:
$$
-18 - \left(\frac{64}{3} - 56\right) = -18 - \frac{64}{3} + 56 = 38 - \frac{64}{3} = \frac{114 - 64}{3} = \frac{50}{3}.
$$

### 6. Total Area

Add the two areas:
$$
\text{Total Area} = \frac{104}{3} + \frac{50}{3} = \frac{154}{3}.
$$
Rounded to the nearest thousandth:
$$
\frac{154}{3} \approx 51.333.
$$

Thus, the total area between $$ f(x) $$ and the $$ x $$-axis from $$ x = 0 $$ to $$ x = 6 $$ is approximately $$ 51.333 $$.
The total area between $$ f(x) = x^2 - x - 12 $$ and the $$ x $$-axis from $$ x = 0 $$ to $$ x = 6 $$ is approximately $$ 51.333 $$.

Find the total area between
and the -axis from to

Total Area ?]
Round your answer to the nearest thousandth.

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Find the total area between and the -axis from to Total Area ?] Round your answer to the nearest thousandth.