To find the area [tex]\( A \)[/tex] of the region bounded by the functions [tex]\( f(x) = \frac{2}{x} \)[/tex] and [tex]\( g(x) = -\frac{x}{5} + \frac{7}{5} \)[/tex], we need to follow these steps:
1. Determine the points of intersection of the curves.
2. Set up the integral to calculate the area between the curves.
### Step 1: Finding the Points of Intersection
We find the points of intersection by setting [tex]\( f(x) \)[/tex] equal to [tex]\( g(x) \)[/tex]:
[tex]\[ \frac{2}{x} = -\frac{x}{5} + \frac{7}{5} \][/tex]
Solving for [tex]\( x \)[/tex], we first clear the fraction by multiplying every term by [tex]\( 5x \)[/tex]:
[tex]\[ 5x \cdot \frac{2}{x} = 5x \cdot \left( -\frac{x}{5} + \frac{7}{5} \right) \][/tex]
[tex]\[ 10 = -x^2 + 7x \][/tex]
Rearrange the equation:
[tex]\[ x^2 - 7x + 10 = 0 \][/tex]
Now solve the quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = 10 \)[/tex]:
[tex]\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{49 - 40}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm 3}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{10}{2} = 5 \][/tex]
[tex]\[ x = \frac{4}{2} = 2 \][/tex]
Therefore, the points of intersection are [tex]\( x = 2 \)[/tex] and [tex]\( x = 5 \)[/tex].
### Step 2: Integrating to Find the Area Between the Curves
The area between two curves from [tex]\( x = a \)[/tex] to [tex]\( x = b \)[/tex] is given by:
[tex]\[ A = \int_{a}^{b} [f(x) - g(x)] \, dx \][/tex]
In our case, [tex]\( a = 2 \)[/tex], [tex]\( b = 5 \)[/tex], [tex]\( f(x) = \frac{2}{x} \)[/tex], and [tex]\( g(x) = -\frac{x}{5} + \frac{7}{5} \)[/tex]:
[tex]\[ A = \int_{2}^{5} \left( \frac{2}{x} - \left( -\frac{x}{5} + \frac{7}{5} \right) \right) \, dx \][/tex]
Simplify the integrand:
[tex]\[ A = \int_{2}^{5} \left( \frac{2}{x} + \frac{x}{5} - \frac{7}{5} \right) \, dx \][/tex]
Integrate each term separately:
[tex]\[ A = \int_{2}^{5} \frac{2}{x} \, dx + \int_{2}^{5} \frac{x}{5} \, dx - \int_{2}^{5} \frac{7}{5} \, dx \][/tex]
Calculate each integral:
[tex]\[ \int_{2}^{5} \frac{2}{x} \, dx = 2 \ln|x| \Big|_{2}^{5} = 2 \ln 5 - 2 \ln 2 = 2 \ln \left( \frac{5}{2} \right) \][/tex]
[tex]\[ \int_{2}^{5} \frac{x}{5} \, dx = \frac{1}{5} \int_{2}^{5} x \, dx = \frac{1}{5} \left( \frac{x^2}{2} \right) \Big|_{2}^{5} = \frac{1}{5} \left( \frac{25}{2} - \frac{4}{2} \right) = \frac{1}{5} \cdot \frac{21}{2} = \frac{21}{10} \][/tex]
[tex]\[ \int_{2}^{5} \frac{7}{5} \, dx = \frac{7}{5} (5 - 2) = \frac{7}{5} \cdot 3 = \frac{21}{5} \][/tex]
Combine the results:
[tex]\[ A = 2 \ln \left( \frac{5}{2} \right) + \frac{21}{10} - \frac{21}{5} \][/tex]
Note that [tex]\( \frac{21}{5} = \frac{42}{10} \)[/tex]:
[tex]\[ A = 2 \ln \left( \frac{5}{2} \right) + \frac{21}{10} - \frac{42}{10} \][/tex]
[tex]\[ A = 2 \ln \left( \frac{5}{2} \right) - \frac{21}{10} \][/tex]
So, the exact area [tex]\( A \)[/tex] of the region [tex]\( R \)[/tex] is:
[tex]\[ A = 2 \ln \left( \frac{5}{2} \right) - \frac{21}{10} \][/tex]
