find the area between the following curves.\nx = - 1, x = 3, y = x^{3}-1, and y = 0\n\nthe area between the…

Explanation:

Step1: Determine the sign of the function

We need to find where (y = x^{3}-1) is above or below (y = 0). Set (x^{3}-1=0), then (x = 1). For (x\in[-1,1]), (x^{3}-1\leqslant0), and for (x\in[1,3]), (x^{3}-1\geqslant0).

Step2: Set up the integral for the area

The area (A=\int_{-1}^{1}(0-(x^{3}-1))dx+\int_{1}^{3}(x^{3}-1 - 0)dx).

Step3: Integrate the first - part

(\int_{-1}^{1}(1 - x^{3})dx=\int_{-1}^{1}1dx-\int_{-1}^{1}x^{3}dx). Since (\int_{-1}^{1}x^{3}dx = 0) (because (y = x^{3}) is an odd function) and (\int_{-1}^{1}1dx=x\big|_{-1}^{1}=1-(-1)=2).

Step4: Integrate the second - part

(\int_{1}^{3}(x^{3}-1)dx=\int_{1}^{3}x^{3}dx-\int_{1}^{3}1dx). We know that (\int x^{3}dx=\frac{1}{4}x^{4}+C) and (\int 1dx=x + C). So (\int_{1}^{3}x^{3}dx=\frac{1}{4}x^{4}\big|{1}^{3}=\frac{81}{4}-\frac{1}{4}=20), and (\int{1}^{3}1dx=x\big|{1}^{3}=3 - 1=2). Then (\int{1}^{3}(x^{3}-1)dx=20 - 2=18).

Step5: Calculate the total area

(A = 2+18=20).

Answer:

20