Answer: Areas of regions Find the area of the following

Question

Problem 41E

Areas of regions

Find the area of the following regions. In each case, graph the relevant functions and show the region in question.

The legion bounded entirely by the curve and the x-axis.

Answer 1

Step-by-Step Solution:

Problem 41E

Find the area of the following regions. In each case, graph the relevant functions and show the region in question.

The region bounded entirely by the curve $y=\frac{{x}^{2}-4x-4}{{x}^{2}-4x-5}$ and the x-axis.
Solution:-
Step 1

The graph of the function $y=\frac{{x}^{2}-4x-4}{{x}^{2}-4x-5}$ is:

The function $y=\frac{{x}^{2}-4x-4}{{x}^{2}-4x-5}$ intersects x-axis at the points when the numerator of the fraction is zero.
Meaning when

${x}^{2}-4x-4=0$
To find the roots of this quadratic equation we see $a=1,b=-4,c=-4$
$D\ ={b}^{2}-4ac$
$D\ ={\left({-4}\right)}^{2}-4\left({1}\right)\left({-4}\right)$
$D=\ 32$
$\sqrt{D}=4\sqrt{2}$

Roots $\left({\alpha{},\beta{}}\right)=\ \frac{-b\pm{}\sqrt{D}}{2a}$

$\alpha{}=\frac{4-4\sqrt{2}}{2},\ \beta{}=\frac{4+4\sqrt{2}}{2}$

$\alpha{}=2-2\sqrt{2},\ \beta{}=2+\sqrt{2}$

As we see from the figure the region bounded by $y=\frac{{x}^{2}-4x-4}{{x}^{2}-4x-5}$ and x-axis is from the point $\left({2-2\sqrt{2},0}\right)\ to\ \left({2+2\sqrt{2},0}\right)$ .

To find the area of the bounded region, we integrate $y=\frac{{x}^{2}-4x-4}{{x}^{2}-4x-5}$ from $\left({x=2-2\sqrt{2}}\right)\ to\ \left({x=2+2\sqrt{2}}\right)$ .

$Area\ =\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\frac{{x}^{2}-4x-4}{{x}^{2}-4x-5}dx\$

= $\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\left({\frac{{x}^{2}-4x-4-1}{{x}^{2}-4x-5}\ +\frac{1}{{x}^{2}-4x-5}}\right)dx$

= $\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\left({\frac{{x}^{2}-4x-5}{{x}^{2}-4x-5}\ +\frac{1}{{x}^{2}-4x-5}}\right)dx$

= $\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}1+\frac{1}{{x}^{2}-5x+x-5}dx$

= $\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\left({1+\frac{1}{x\left({x-5}\right)+1\left({x-5}\right)}}\right)dx$
= $\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\left({1+\frac{1}{\left({x+1}\right)\left({x-5}\right)}}\right)dx$

= $x\ +\ \left({\frac{-1}{6}}\right)\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\frac{-6}{\left({x+1}\right)\left({x-5}\right)}dx$

= $x\ +\ \left({\frac{-1}{6}}\right)\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\frac{\left({x-5}\right)-\left({x+1}\right)}{\left({x+1}\right)\left({x-5}\right)}dx$
= $x\ +\ \left({\frac{-1}{6}}\right)\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\left({\frac{1}{\left({x+1}\right)}+\frac{-1}{\left({x-5}\right)}}\right)dx$
= $x\ +\ \left({\frac{-1}{6}}\right)\ \int\limits_{2-2\sqrt{2}}^{2+2\sqrt{2}}\left({\frac{1}{\left({x+1}\right)}+\frac{1}{\left({5-x}\right)}}\right)dx$
= $x\ +$ $\ \left({-\frac{1}{6}}\right)$ $\left({ln\left({x+1}\right)-ln\left({5-x}\right)}\right)$
= $x\ +\frac{1}{6}ln\left({5-x}\right)-\frac{1}{6}ln\left({x+1}\right)$

Putting the limits in we get the area, we get

$I\ =\ 4{\sqrt{2}}_{\ }-\frac{1}{3}ln\left({3+2\sqrt{2}}\right)+\frac{1}{3}ln\left({3-2\sqrt{2}}\right)=4.48$

Step 2 of 1