Find the total area between the $$ x $$-axis and the function on the given interval. $$ f(x)=x^{\frac{1}{3}}-x $$ $$ [-1,8] $$ $$ [?] $$

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We are given 
$$
f(x)=x^{\frac{1}{3}}-x
$$
and we wish to find the total area between the $$ x $$-axis and $$ f(x) $$ on the interval $$[-1,8]$$. Since the total area involves the absolute value of the function, we must first locate the points where $$ f(x)=0 $$ on the interval.

**1. Find the zeros of $$ f(x) $$:**

We solve
$$
x^{\frac{1}{3}}-x=0 \quad \Longrightarrow \quad x^{\frac{1}{3}}=x.
$$
Cubing both sides (using the property $$\left(x^{\frac{1}{3}}\right)^3 = x$$) we obtain
$$
x = x^3.
$$
Rearranging gives
$$
x^3-x=0 \quad \Longrightarrow \quad x(x^2-1)=0.
$$
Thus,
$$
x=0 \quad \text{or} \quad x^2-1=0 \quad \Longrightarrow \quad x=\pm1.
$$
On the interval $$[-1,8]$$ the zeros are $$ x=-1 $$, $$ x=0 $$, and $$ x=1 $$.

**2. Determine the sign of $$ f(x) $$ on the subintervals:**

- On $$[-1,0]$$: Choose $$ x=-0.5 $$. Then
  $$
  f(-0.5)=(-0.5)^{\frac{1}{3}}-(-0.5).
  $$
  Since the cube root of a negative number is negative, $$ (-0.5)^{\frac{1}{3}} $$ is negative. A quick estimate shows $$ f(-0.5) $$ is negative.

- On $$[0,1]$$: Choose $$ x=0.5 $$. Then
  $$
  f(0.5)=0.5^{\frac{1}{3}}-0.5.
  $$
  This value turns out to be positive.

- On $$[1,8]$$: For example, at $$ x=2 $$ or $$ x=8 $$, $$ f(x) $$ is negative.

Thus, the function is negative on $$[-1,0]$$ and $$[1,8]$$ and positive on $$[0,1]$$.

**3. Set up the integrals for the total (absolute) area:**

The total area $$ A $$ is given by
$$
A = \left| \int_{-1}^{0} f(x)\,dx \right| + \left| \int_{0}^{1} f(x)\,dx \right| + \left| \int_{1}^{8} f(x)\,dx \right|.
$$

**4. Find the antiderivative of $$ f(x) $$:**

We write
$$
\int f(x)\,dx = \int \left(x^{\frac{1}{3}}-x\right)dx = \int x^{\frac{1}{3}}dx-\int x\,dx.
$$
The antiderivatives are:
$$
\int x^{\frac{1}{3}}dx = \frac{3}{4}x^{\frac{4}{3}} \quad \text{and} \quad \int x\,dx = \frac{x^2}{2}.
$$
Thus, the antiderivative $$ F(x) $$ is
$$
F(x)=\frac{3}{4}x^{\frac{4}{3}}-\frac{x^2}{2}.
$$

**5. Evaluate the integrals on each subinterval:**

- **Interval $$[-1,0]$$:**
  $$
  I_1=\int_{-1}^{0} \left(x^{\frac{1}{3}}-x\right)dx = F(0)-F(-1).
  $$
  Compute:
  $$
  F(0)=\frac{3}{4}\cdot0^{\frac{4}{3}}-\frac{0^2}{2}=0,
  $$
  and note that for $$ x=-1 $$,
  $$
  (-1)^{\frac{1}{3}}=-1 \quad \text{and} \quad (-1)^{\frac{4}{3}}=\left[(-1)^{\frac{1}{3}}\right]^4= (-1)^4=1.
  $$
  Therefore,
  $$
  F(-1)=\frac{3}{4}\cdot 1-\frac{1}{2}=\frac{3}{4}-\frac{1}{2}=\frac{3-2}{4}=\frac{1}{4}.
  $$
  So,
  $$
  I_1=0-\frac{1}{4}=-\frac{1}{4}.
  $$
  Since $$ f(x) $$ is negative on this interval, the area is
  $$
  A_1=\left| I_1 \right|=\frac{1}{4}.
  $$

- **Interval $$[0,1]$$:**
  $$
  I_2=\int_{0}^{1} \left(x^{\frac{1}{3}}-x\right)dx = F(1)-F(0).
  $$
  We already have $$ F(0)=0 $$. For $$ x=1 $$,
  $$
  1^{\frac{4}{3}}=1,
  $$
  so
  $$
  F(1)=\frac{3}{4}\cdot 1-\frac{1}{2}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}.
  $$
  The area on $$[0,1]$$ is
  $$
  A_2=I_2=\frac{1}{4}.
  $$

- **Interval $$[1,8]$$:**
  $$
  I_3=\int_{1}^{8} \left(x^{\frac{1}{3}}-x\right)dx = F(8)-F(1).
  $$
  First, compute $$ F(8) $$. Notice that
  $$
  8^{\frac{1}{3}}=2 \quad \Longrightarrow \quad 8^{\frac{4}{3}}=(8^{\frac{1}{3}})^4=2^4=16.
  $$
  Then,
  $$
  F(8)=\frac{3}{4}\cdot16-\frac{8^2}{2} = 12- \frac{64}{2} =12-32=-20.
  $$
  We have already found $$ F(1)=\frac{1}{4} $$. So,
  $$
  I_3=-20-\frac{1}{4}=-\frac{80}{4}-\frac{1}{4}=-\frac{81}{4}.
  $$
  Since $$ f(x) $$ is negative on $$[1,8]$$, the area is
  $$
  A_3=\left| I_3 \right|=\frac{81}{4}.
  $$

**6. Sum the areas:**

The total area is
$$
A=A_1+A_2+A_3=\frac{1}{4}+\frac{1}{4}+\frac{81}{4}=\frac{1+1+81}{4}=\frac{83}{4}.
$$

Thus, the total area between the $$ x $$-axis and the curve $$ f(x)=x^{\frac{1}{3}}-x $$ on the interval $$[-1,8]$$ is
$$
\boxed{\frac{83}{4}}.
$$
The total area between the $$ x $$-axis and the function $$ f(x) = x^{\frac{1}{3}} - x $$ on the interval $$[-1,8]$$ is $$ \frac{83}{4} $$.

Find the total area between the \( x \)-axis and the function on the given interval. \[ f(x)=x^{\frac{1}{3}}-x \] \( [-1,8] \) \( [?] \)

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