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

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 \[ \boxed{\frac{83}{4}}. \]