Math Problem Statement

Find the area A of the region that is bounded between the curve f(x)=ln(x)+1 and the line g(x)=−4x/e +6 over the interval [1,4].

Solution

To find the area AA of the region bounded between the curve f(x)=ln(x)+1f(x) = \ln(x) + 1 and the line g(x)=4xe+6g(x) = -\frac{4x}{e} + 6 over the interval [1,4][1, 4], we need to perform the following steps:

1. Determine the points of intersection

The points of intersection occur where f(x)=g(x)f(x) = g(x). Set the two equations equal to each other and solve for xx:

ln(x)+1=4xe+6\ln(x) + 1 = -\frac{4x}{e} + 6

This equation is transcendental, so finding the exact solutions analytically is challenging. However, we can approximate the intersection points numerically or confirm that the intersection points occur at the boundaries x=1x = 1 and x=4x = 4.

2. Set up the definite integral

The area between the two curves can be calculated using the integral:

A=14[g(x)f(x)]dxA = \int_{1}^{4} \left[ g(x) - f(x) \right] \, dx

Substitute the expressions for f(x)f(x) and g(x)g(x):

A=14((4xe+6)(ln(x)+1))dxA = \int_{1}^{4} \left( \left(-\frac{4x}{e} + 6\right) - \left(\ln(x) + 1\right) \right) \, dx

Simplify the integrand:

A=14(4xe+5ln(x))dxA = \int_{1}^{4} \left(-\frac{4x}{e} + 5 - \ln(x)\right) \, dx

3. Evaluate the integral

Now, we can integrate term by term:

144xedx=4e14xdx=4e[x22]14=4e(16212)=4e152=30e\int_{1}^{4} -\frac{4x}{e} \, dx = -\frac{4}{e} \int_{1}^{4} x \, dx = -\frac{4}{e} \left[\frac{x^2}{2}\right]_{1}^{4} = -\frac{4}{e} \left(\frac{16}{2} - \frac{1}{2}\right) = -\frac{4}{e} \cdot \frac{15}{2} = -\frac{30}{e}

145dx=5141dx=5[x]14=5(41)=15\int_{1}^{4} 5 \, dx = 5 \int_{1}^{4} 1 \, dx = 5[x]_{1}^{4} = 5(4 - 1) = 15

14ln(x)dx=[xln(x)x]14=(4ln(4)4)(1ln(1)1)=4ln(4)4+1=4ln(4)3\int_{1}^{4} \ln(x) \, dx = \left[x\ln(x) - x\right]_{1}^{4} = \left(4\ln(4) - 4\right) - (1\ln(1) - 1) = 4\ln(4) - 4 + 1 = 4\ln(4) - 3

Thus, the area AA is:

A=30e+15(4ln(4)3)=1830e4ln(4)A = -\frac{30}{e} + 15 - (4\ln(4) - 3) = 18 - \frac{30}{e} - 4\ln(4)

Final answer:

A=1830e4ln(4)A = 18 - \frac{30}{e} - 4\ln(4)

This is the exact expression for the area AA of the region bounded between the curve and the line.

Would you like further details or have any questions?

Related Questions:

  1. How would the result change if the interval were different?
  2. Can we approximate this area using numerical methods?
  3. What is the geometric interpretation of the integral used here?
  4. How would the area calculation differ if the line equation were steeper?
  5. Can this method be applied to other types of curves, such as quadratic functions?

Tip:

When dealing with integrals that involve logarithmic functions, remember that integration by parts might sometimes be necessary if the integral doesn't directly resolve.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Logarithmic Functions

Formulas

Area between two curves: A = ∫[a,b] (g(x) - f(x)) dx
Integration of logarithmic functions: ∫ ln(x) dx = x ln(x) - x

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 or College Level