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.

Question

UpStudy AI · Accepted Answer

We first find where the function
$$
f(x)=x^2-x-12
$$
crosses the $$ x $$-axis by solving
$$
x^2 - x - 12 = 0.
$$

### Step 1. Find the Roots

Factor the quadratic:
$$
x^2 - x - 12 = (x-4)(x+3)=0.
$$
Thus, the roots are:
$$
x=4 \quad \text{and} \quad x=-3.
$$

Since we are interested in the interval from $$ x=0 $$ to $$ x=6 $$, only $$ x=4 $$ is relevant.

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

Evaluate the function at a point between $$ x=0 $$ and $$ x=4 $$, for example, $$ x=0 $$:
$$
f(0)=0^2-0-12=-12.
$$
Since $$ f(0) < 0 $$, the function is below the $$ x $$-axis on the interval from $$ x=0 $$ to $$ x=4 $$.

Next, evaluate at a point between $$ x=4 $$ and $$ x=6 $$, for example, $$ x=6 $$:
$$
f(6)=6^2-6-12=36-6-12=18.
$$
This tells us that $$ f(x) $$ is above the $$ x $$-axis from $$ x=4 $$ to $$ x=6 $$.

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

The total area between the curve and the $$ x $$-axis is the sum of the area below the axis from $$ x=0 $$ to $$ x=4 $$ (taking the absolute value) and the area above the axis from $$ x=4 $$ to $$ x=6 $$.

The antiderivative of $$ f(x) $$ is:
$$
F(x)=\int (x^2-x-12)\,dx=\frac{x^3}{3}-\frac{x^2}{2}-12x.
$$

#### Area from $$ x=0 $$ to $$ x=4 $$:
Since $$ f(x) $$ is negative on this interval,
$$
\text{Area}_1 = -\int_{0}^{4}(x^2-x-12)\,dx.
$$

Compute $$ F(4) $$:
$$
F(4)=\frac{4^3}{3}-\frac{4^2}{2}-12(4)=\frac{64}{3}-\frac{16}{2}-48=\frac{64}{3}-8-48=\frac{64}{3}-56.
$$
Compute $$ F(0) $$:
$$
F(0)=0.
$$
Thus,
$$
\int_{0}^{4}(x^2-x-12)\,dx = F(4)-F(0)=\frac{64}{3}-56.
$$
Since this value is negative, the area is:
$$
\text{Area}_1 = -\left(\frac{64}{3}-56\right)=56-\frac{64}{3}=\frac{168}{3}-\frac{64}{3}=\frac{104}{3}.
$$

#### Area from $$ x=4 $$ to $$ x=6 $$:
On this interval, $$ f(x) $$ is positive, so the area is:
$$
\text{Area}_2 = \int_{4}^{6}(x^2-x-12)\,dx.
$$

Compute $$ F(6) $$:
$$
F(6)=\frac{6^3}{3}-\frac{6^2}{2}-12(6)=\frac{216}{3}-\frac{36}{2}-72=72-18-72=-18.
$$
Using the previously computed $$ F(4) $$:
$$
F(4)=\frac{64}{3}-56.
$$
Thus,
$$
\int_{4}^{6}(x^2-x-12)\,dx = F(6)-F(4)=-18-\left(\frac{64}{3}-56\right)=-18-\frac{64}{3}+56.
$$
Simplify:
$$
-18 + 56 = 38,
$$
so,
$$
\text{Area}_2 = 38-\frac{64}{3}=\frac{114}{3}-\frac{64}{3}=\frac{50}{3}.
$$

### Step 4. Compute the Total Area

The total area is:
$$
\text{Total Area} = \text{Area}_1+\text{Area}_2 = \frac{104}{3}+\frac{50}{3}=\frac{154}{3}.
$$

Convert to a decimal:
$$
\frac{154}{3} \approx 51.333.
$$

### Final Answer

$$
\boxed{51.333}
$$
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.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Step 1. Find the Roots

Step 2. Determine the Sign of

Step 3. Set Up the Integrals for Total Area

Area from to :

Area from to :

Step 4. Compute the Total Area

Final Answer

Mind Expander

preguntas relacionadas

Latest Calculus Questions

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

Solución de inteligencia artificial de Upstudy

Responder

Solución

Step 1. Find the Roots

Step 2. Determine the Sign of

Step 3. Set Up the Integrals for Total Area

Area from to :

Area from to :

Step 4. Compute the Total Area

Final Answer

Mind Expander

preguntas relacionadas

Latest Calculus Questions

Find the total area between
and the -axis from to

Total Area
Round your answer to the nearest thousandth.