5) Find the area of the region bounded by the graphs of $$ f(x)=x^{3}-x^{2}-10 x $$ and $$ g(x)=-x^{2}+2 x $$

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1. **Find the Points of Intersection**

Set 
   $$
   f(x)=x^3-x^2-10x \quad \text{and} \quad g(x)=-x^2+2x
   $$
   equal to each other:
   $$
   x^3-x^2-10x=-x^2+2x.
   $$
   
   Bring all terms to one side:
   $$
   x^3-x^2-10x + x^2 - 2x = 0.
   $$
   
   Combine like terms:
   $$
   x^3 -12x = 0.
   $$
   
   Factor:
   $$
   x(x^2-12)=0.
   $$
   
   The solutions are:
   $$
   x=0,\quad x^2-12=0 \implies x=\pm 2\sqrt{3}.
   $$
   
   So, the intersection points occur at $$ x=-2\sqrt{3} $$, $$ x=0 $$, and $$ x=2\sqrt{3} $$.

2. **Determine Which Function is on Top**

Compute the difference:
   $$
   f(x)-g(x) = \left(x^3-x^2-10x\right)-\left(-x^2+2x\right)=x^3-12x.
   $$
   
   - For an $$ x $$ in the interval $$ (-2\sqrt{3}, 0) $$, take $$ x=-1 $$:
     $$
     f(-1)-g(-1)=(-1)^3-12(-1)=-1+12=11>0.
     $$
     This implies $$ f(x)>g(x) $$ on $$ (-2\sqrt{3},0) $$.
     
   - For an $$ x $$ in the interval $$ (0, 2\sqrt{3}) $$, take $$ x=1 $$:
     $$
     f(1)-g(1)=1^3-12(1)=1-12=-11<0.
     $$
     This implies $$ g(x)>f(x) $$ on $$ (0, 2\sqrt{3}) $$.

3. **Set Up the Integral for the Area**

The area between the curves is given by the sum of the areas in the two intervals.

- On $$ \left[-2\sqrt{3},0\right] $$, since $$ f(x) $$ is above $$ g(x) $$:
     $$
     A_1=\int_{-2\sqrt{3}}^{0}\left[f(x)-g(x)\right]dx=\int_{-2\sqrt{3}}^{0}\left(x^3-12x\right)dx.
     $$
     
   - On $$ \left[0,2\sqrt{3}\right] $$, since $$ g(x) $$ is above $$ f(x) $$:
     $$
     A_2=\int_{0}^{2\sqrt{3}}\left[g(x)-f(x)\right]dx=\int_{0}^{2\sqrt{3}}\left(-x^3+12x\right)dx.
     $$
     
   The total area is:
   $$
   A=A_1+A_2.
   $$

4. **Evaluate the Integrals**

**For $$ A_1 $$:**
   
   Find the antiderivative:
   $$
   \int \left(x^3-12x\right)dx=\frac{x^4}{4}-6x^2.
   $$
   
   Evaluate from $$ x=-2\sqrt{3} $$ to $$ x=0 $$:
   $$
   A_1=\left[\frac{x^4}{4}-6x^2\right]_{-2\sqrt{3}}^{0}=\left(\frac{0^4}{4}-6\cdot0^2\right)-\left(\frac{(-2\sqrt{3})^4}{4}-6(-2\sqrt{3})^2\right).
   $$
   
   Note that:
   $$
   (-2\sqrt{3})^2=4\cdot3=12 \quad \text{and} \quad (-2\sqrt{3})^4=(12)^2=144.
   $$
   
   So,
   $$
   \frac{144}{4}-6\cdot12=36-72=-36.
   $$
   
   Hence,
   $$
   A_1=0-(-36)=36.
   $$
   
   **For $$ A_2 $$:**
   
   Find the antiderivative:
   $$
   \int\left(-x^3+12x\right)dx=-\frac{x^4}{4}+6x^2.
   $$
   
   Evaluate from $$ x=0 $$ to $$ x=2\sqrt{3} $$:
   $$
   A_2=\left[-\frac{x^4}{4}+6x^2\right]_{0}^{2\sqrt{3}}=\left(-\frac{(2\sqrt{3})^4}{4}+6(2\sqrt{3})^2\right)-0.
   $$
   
   With:
   $$
   (2\sqrt{3})^2=12 \quad \text{and} \quad (2\sqrt{3})^4=144,
   $$
   
   we get:
   $$
   -\frac{144}{4}+6\cdot12=-36+72=36.
   $$
   
5. **Calculate the Total Area**

$$
   A=A_1+A_2=36+36=72.
   $$

Thus, the area of the region bounded by the curves is $$72$$.
The area of the region bounded by the graphs of $$ f(x)=x^{3}-x^{2}-10x $$ and $$ g(x)=-x^{2}+2x $$ is 72.

Find the area of the region bounded by the graphs of
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