Answer :
To solve for the area [tex]\( A \)[/tex] of the region [tex]\( R \)[/tex] bounded by the functions [tex]\( f(x) = -\frac{4}{x} \)[/tex] and [tex]\( g(x) = \frac{2x}{7} - \frac{18}{7} \)[/tex], follow these steps:
1. Find the Points of Intersection:
To determine the intersection points of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], set [tex]\( f(x) \)[/tex] equal to [tex]\( g(x) \)[/tex]:
[tex]\[ -\frac{4}{x} = \frac{2x}{7} - \frac{18}{7} \][/tex]
Solving this equation involves clearing the fractions and solving for [tex]\( x \)[/tex]:
[tex]\[ -4 = \left(\frac{2x}{7} - \frac{18}{7}\right)x \][/tex]
Multiplying both sides by 7 to clear the denominators:
[tex]\[ -28 = 2x^2 - 18x \][/tex]
Now, rearrange this into a standard quadratic equation:
[tex]\[ 2x^2 - 18x + 28 = 0 \][/tex]
Simplifying by dividing the entire equation by 2:
[tex]\[ x^2 - 9x + 14 = 0 \][/tex]
Now factorize:
[tex]\[ (x - 2)(x - 7) = 0 \][/tex]
Solving for [tex]\( x \)[/tex] gives the points of intersection:
[tex]\[ x = 2 \quad \text{or} \quad x = 7 \][/tex]
2. Set Up the Integral:
To find the area between two curves, subtract the lower function from the upper function and integrate over the interval from 2 to 7:
[tex]\[ A = \int_{2}^{7} \left(g(x) - f(x)\right) \, dx \][/tex]
Substitute the functions [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex]:
[tex]\[ A = \int_{2}^{7} \left(\left(\frac{2x}{7} - \frac{18}{7}\right) - \left(-\frac{4}{x}\right)\right) \, dx \][/tex]
Simplify the integrand:
[tex]\[ A = \int_{2}^{7} \left(\frac{2x}{7} - \frac{18}{7} + \frac{4}{x}\right) \, dx \][/tex]
3. Evaluate the Integral:
Separate the integrals:
[tex]\[ A = \int_{2}^{7} \frac{2x}{7} \, dx - \int_{2}^{7} \frac{18}{7} \, dx + \int_{2}^{7} \frac{4}{x} \, dx \][/tex]
Evaluate each integral:
[tex]\[ \int_{2}^{7} \frac{2x}{7} \, dx = \frac{2}{7} \int_{2}^{7} x \, dx = \frac{2}{7} \left[ \frac{x^2}{2} \right]_{2}^{7} = \frac{2}{7} \left( \frac{49}{2} - \frac{4}{2} \right) = \frac{2}{7} \cdot \frac{45}{2} = \frac{45}{7} \][/tex]
[tex]\[ \int_{2}^{7} \frac{18}{7} \, dx = \frac{18}{7} \int_{2}^{7} 1 \, dx = \frac{18}{7} \left[ x \right]_{2}^{7} = \frac{18}{7} (7 - 2) = \frac{18}{7} \cdot 5 = \frac{90}{7} \][/tex]
[tex]\[ \int_{2}^{7} \frac{4}{x} \, dx = 4 \int_{2}^{7} \frac{1}{x} \, dx = 4 \left[ \ln |x| \right]_{2}^{7} = 4 (\ln 7 - \ln 2) = 4 \ln \left( \frac{7}{2} \right) \][/tex]
Combine the evaluated integrals:
[tex]\[ A = \frac{45}{7} - \frac{90}{7} + 4 \ln \left( \frac{7}{2} \right) = -\frac{45}{7} + 4 \ln \left( \frac{7}{2} \right) \][/tex]
Given the numeric result:
A = -1.4175195545899566，通过
Therefore, the area [tex]\( A \)[/tex] of the region is:
[tex]\[ A = -1.4175195545899566 \text { units}^2 \][/tex]
1. Find the Points of Intersection:
To determine the intersection points of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], set [tex]\( f(x) \)[/tex] equal to [tex]\( g(x) \)[/tex]:
[tex]\[ -\frac{4}{x} = \frac{2x}{7} - \frac{18}{7} \][/tex]
Solving this equation involves clearing the fractions and solving for [tex]\( x \)[/tex]:
[tex]\[ -4 = \left(\frac{2x}{7} - \frac{18}{7}\right)x \][/tex]
Multiplying both sides by 7 to clear the denominators:
[tex]\[ -28 = 2x^2 - 18x \][/tex]
Now, rearrange this into a standard quadratic equation:
[tex]\[ 2x^2 - 18x + 28 = 0 \][/tex]
Simplifying by dividing the entire equation by 2:
[tex]\[ x^2 - 9x + 14 = 0 \][/tex]
Now factorize:
[tex]\[ (x - 2)(x - 7) = 0 \][/tex]
Solving for [tex]\( x \)[/tex] gives the points of intersection:
[tex]\[ x = 2 \quad \text{or} \quad x = 7 \][/tex]
2. Set Up the Integral:
To find the area between two curves, subtract the lower function from the upper function and integrate over the interval from 2 to 7:
[tex]\[ A = \int_{2}^{7} \left(g(x) - f(x)\right) \, dx \][/tex]
Substitute the functions [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex]:
[tex]\[ A = \int_{2}^{7} \left(\left(\frac{2x}{7} - \frac{18}{7}\right) - \left(-\frac{4}{x}\right)\right) \, dx \][/tex]
Simplify the integrand:
[tex]\[ A = \int_{2}^{7} \left(\frac{2x}{7} - \frac{18}{7} + \frac{4}{x}\right) \, dx \][/tex]
3. Evaluate the Integral:
Separate the integrals:
[tex]\[ A = \int_{2}^{7} \frac{2x}{7} \, dx - \int_{2}^{7} \frac{18}{7} \, dx + \int_{2}^{7} \frac{4}{x} \, dx \][/tex]
Evaluate each integral:
[tex]\[ \int_{2}^{7} \frac{2x}{7} \, dx = \frac{2}{7} \int_{2}^{7} x \, dx = \frac{2}{7} \left[ \frac{x^2}{2} \right]_{2}^{7} = \frac{2}{7} \left( \frac{49}{2} - \frac{4}{2} \right) = \frac{2}{7} \cdot \frac{45}{2} = \frac{45}{7} \][/tex]
[tex]\[ \int_{2}^{7} \frac{18}{7} \, dx = \frac{18}{7} \int_{2}^{7} 1 \, dx = \frac{18}{7} \left[ x \right]_{2}^{7} = \frac{18}{7} (7 - 2) = \frac{18}{7} \cdot 5 = \frac{90}{7} \][/tex]
[tex]\[ \int_{2}^{7} \frac{4}{x} \, dx = 4 \int_{2}^{7} \frac{1}{x} \, dx = 4 \left[ \ln |x| \right]_{2}^{7} = 4 (\ln 7 - \ln 2) = 4 \ln \left( \frac{7}{2} \right) \][/tex]
Combine the evaluated integrals:
[tex]\[ A = \frac{45}{7} - \frac{90}{7} + 4 \ln \left( \frac{7}{2} \right) = -\frac{45}{7} + 4 \ln \left( \frac{7}{2} \right) \][/tex]
Given the numeric result:
A = -1.4175195545899566，通过
Therefore, the area [tex]\( A \)[/tex] of the region is:
[tex]\[ A = -1.4175195545899566 \text { units}^2 \][/tex]
