Answer
$\frac{83}{4}$
Work Step by Step
Step 1. Given the function $y=x^{1/3}-x=x^{1/3}(1-x^{2/3})$ and the interval $[-1,8]$, find the zeros at $x=0,1$; thus we need to divide the integration into three regions: $[-1,0],[0,1],[1,8]$
Step 2. We have $A_1=\int_{-1}^{0}(x^{1/3}-x)dx=(\frac{3}{4}x^{4/3}-\frac{1}{2}x^2)|_{-1}^{0}=(\frac{3}{4}(0)^{4/3}-\frac{1}{2}(0)^2)-(\frac{3}{4}(-1)^{4/3}-\frac{1}{2}(-1)^2)=-\frac{1}{4}$
Step 3. Similarly, $A_2=\int_{0}^{1}(x^{1/3}-x)dx=(\frac{3}{4}x^{4/3}-\frac{1}{2}x^2)|_{0}^{1}=(\frac{3}{4}(1)^{4/3}-\frac{1}{2}(1)^2)-(\frac{3}{4}(0)^{4/3}-\frac{1}{2}(0)^2)=\frac{1}{4}$
Step 4. For the third part, $A_3=\int_{1}^{8}(x^{1/3}-x)dx=(\frac{3}{4}x^{4/3}-\frac{1}{2}x^2)|_{1}^{8}=(\frac{3}{4}(8)^{4/3}-\frac{1}{2}(8)^2)-(\frac{3}{4}(1)^{4/3}-\frac{1}{2}(1)^2)=\frac{81}{4}$
Step 5. The total area is given by $A=|A_1|+|A_2+|A_3|=\frac{83}{4}$
