Chapter 6.4, Problem 43E

To determine

Total area of the region between the curve and x-axis.

Answer to Problem 43E

Between the curve and x-axis,

  $\text{Total}\text{area}=\frac{1}{2}$

Explanation of Solution

Given information:

  $y=x^3-3x^2+2x$

Such that

  $0\le x\le 2$

We are required to find the total area.

That means

Any area below the x-axis needs to be counted as a positive area.

Graph the function

  $f\left(x\right)=x^3-3x^2+2x$

On the interval [0, 2], to see the function which is above and below the x-axis.

  

On the interval [0, 1]:

  ${\displaystyle {\int }_0^1\left(x^3-3x^2+2x\right)dx}={\left(\frac{1}{4}{x}^4-x^3+x^2\right)|}_0^1=\left(\frac{1}{4}-1+1\right)-\left(0-0+0\right)=\frac{1}{4}-0=\frac{1}{4}$

On the interval [1, 2]:

  ${\displaystyle {\int }_1^2\left(x^3-3x^2+2x\right)dx}={\left(\frac{1}{4}{x}^4-x^3+x^2\right)|}_1^2=\left(\frac{1}{4}{\left(2\right)}^4-{\left(2\right)}^3+{\left(2\right)}^2\right)-\left(\frac{1}{4}-1+1\right)=0-\frac{1}{4}=-\frac{1}{4}$

Thus,

  $\text{Total}\text{area}=\left|\frac{1}{4}\right|+\left|-\frac{1}{4}\right|=\frac{1}{2}$