Question

Each of the definite integrals in Exercises 19–24 represents the volume of a solid of revolution obtained by rotating a region around either the x- or y-axis. Find this region.

$\mathrm{\pi }{\int }_1^3({\mathrm{x}}^2-2\mathrm{x}+1)\mathrm{dx}$

Short answer

The required region is shown as,

Step-by-step solution

Step 1. Given Information  

The given integral is $\mathrm{\pi }{\int }_1^3({\mathrm{x}}^2-2\mathrm{x}+1)\mathrm{dx}$

Step 2. Explanation  

The given integral is of the form $\mathrm{\pi }{\int }_a^b(\mathrm{f}{\left(\mathrm{x}\right))}^2\mathrm{dx}$ which represents the volume of a solid formed by rotating the region bound by f(x) and x-axis in the interval $[a,b]$ around x-axis.

On comparing the given region with the above region, we have,

$f\left(x\right)=\sqrt{x^2-2x+1}\mathrm{in}\mathrm{the}\mathrm{interval}[1,3]$

Factor the quadratic expression inside the radical sign to simplify it

$$\begin{gathered} f\left(x\right)=\sqrt{x^2-2x+1} \\ =\sqrt{{(x-1)}^2} \\ =x-1 \end{gathered}$$

The simplified function represents the straight line.

Plot the graph of the function and identify the region between the intervals.