Chapter 6.1, Problem 7E

To determine

To draw: the region enclosed by given curve.

The area of the region enclosed by the curves.

Answer to Problem 7E

The area of the region enclosed by the curves is $\underset{\_}{\frac{9}{2}}$.

Explanation of Solution

Given information:

The two curves has a function of $y=x$ and $y={\left(x-2\right)}^2$.

Calculation:

Find the intersection points of the curves by equating the curves as shown below:

$f\left(x\right)=g\left(x\right)$ (1)

Here, the top curve function is $f\left(x\right)$ and the bottom curve function is $g\left(x\right)$.

Substitute x for $f\left(x\right)$ and ${\left(x-2\right)}^2$ for $g\left(x\right)$ in Equation (1).

$\begin{array}{l}x={\left(x-2\right)}^2\\ x=x^2-4x+4\\ x^2-4x-x+4=0\\ x^2-5x+4=0\end{array}$ (2)

Solve Equation (2).

$\begin{array}{l}{x}^2-5x+4=0\\ x^2-1x-4x+4=0\\ x\left(x-1\right)-4\left(x-1\right)\\ x=1\text{and}4\end{array}$

Procedure to sketch the region bounded by the two curves is explained below:

• Draw the graph for the function $y={\left(x-2\right)}^2$ by substituting different values for x.
• Similarly in the same graph plot for the function $y=x$ by substituting different values for x.
• Shade the region lies between $x=1$ and $x=4$.

The region enclosed by the curves $y={\left(x-2\right)}^2$ and $y=x$ is shown in Figure 1.

Decide whether the integration can be done with respect to x or y is given below:

• If the region of curves is bound by the top and bottom curve then the integration can be done with respect to x.
• If the region of curves is bound by the right and left curve then the integration can be done with respect to y.

Therefore, the integration can be done with respect to x.

Refer to Figure (1).

Draw a typical approximate rectangle with a width $\left(\Delta x\right)$ and height $\left(h\right)$.

Find the height of the typical approximating rectangle using the relation:

$h=y_T-y_B$ (3)

Here, the top curve is $y_T$ and the bottom curve is $y_B$.

Substitute $x$  for $y_T$ and ${\left(x-2\right)}^2$ for $y_B$ in Equation (3).

$h=x-{\left(x-2\right)}^2$

The typical approximating rectangle with dimensions for the region bounded by the curves is shown in Figure 2.

Find the area of the region bounded by the curves using the relation:

$A={\displaystyle {\int }_a^b\left(f\left(x\right)-g\left(x\right)\right)}dx$ (4)

Here, the lower limit is a and the upper limit is b.

Substitute $x$ for $f\left(x\right)$, ${\left(x-2\right)}^2$ for $g\left(x\right)$, $1$ for a, and $4$ for b in Equation (4).

$\begin{array}{c}A={\displaystyle {\int }_1^4\left[x-{\left(x-2\right)}^2\right]}dx\\ ={\displaystyle {\int }_1^4\left[x-\left(x^2-4x+4\right)\right]}dx\\ ={\displaystyle {\int }_1^4\left[-x^2+5x-4\right]}dx\\ ={\left[-\frac{x^3}{3}+5\frac{x^2}{2}-4x\right]}_1^4\end{array}$

$\begin{array}{c}=\left(-\frac{4^3}{3}+5\frac{4^2}{2}-4\left(4\right)\right)-\left(-\frac{1^3}{3}+5\frac{1^2}{2}-4\left(1\right)\right)\\ =-\frac{64}{3}+\frac{80}{2}-16+\frac{1}{3}-\frac{5}{2}+4\\ =\frac{-128+240-96+2-15+24}{6}\\ =\frac{27}{6}\end{array}$

$=\frac{9}{2}$

Therefore, the area of the region enclosed by the curves is $\underset{\_}{\frac{9}{2}}$.