Question

Set up the definite integral that gives the area of the region. $$ \begin{array}{l} f(x)=(x-1)^{3} \\ g(x)=x-1 \end{array} $$

Short answer

The definite integral that gives the area of the region between the two curves \(f(x)=(x-1)^3\) and \(g(x)=x-1\) from \(x=0\) to \(x=1\) is \(\int_{0}^{1} ( (x-1)^{3} - (x-1) ) dx\).

Step-by-step solution

Finding Points of Intersection

Set f(x) equal to g(x) to find the points where the two graphs intersect. So, set \((x-1)^{3}\) equal to \(x-1\). Solve for \(x\). This gives \(x=1, 0\).

Determining the Order of Functions

Between \(x=0\) and \(x=1\), check which function is above the other. For example, plug in \(x = 0.5\) in both f(x) and g(x). Here, f(x) gives \(0.5 - 1 = -0.5\), and g(x) gives \((0.5-1)^{3} = -0.125\). Since \(g(x) > f(x)\) for \(x = 0.5\), in this range, the curve \(f(x)=(x-1)^{3}\) is above \(g(x)=x-1\).

Integral Setup for Area Calculation

Setup the definite integral for the area between the curves from \(x = 0\) to \(x = 1\), defined as the absolute difference of the two functions. So, the integral is \(\int_{0}^{1}\mid f(x)-g(x) \mid dx = \int_{0}^{1} ( (x-1)^{3} - (x-1) ) dx\).