Calculate The Area Between F(x) = Ln(x) + 1 And G(x) = -4x/e + 6

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Finding the area of a region bounded by curves is a fundamental concept in calculus, with applications in various fields such as physics, engineering, and economics. This article delves into the process of calculating the area A of the region enclosed between the curve f(x) = ln(x) + 1 and the line g(x) = -4x/e + 6 over the interval [1, 4]. We will explore the necessary steps, including setting up the integral and evaluating it to arrive at the exact answer. This comprehensive guide will provide a clear understanding of the methods involved and the reasoning behind them.

Understanding the Problem

The problem asks us to determine the area A of the region that lies between two functions, f(x) = ln(x) + 1 and g(x) = -4x/e + 6, within the interval [1, 4]. To visualize this, imagine graphing both functions on the same coordinate plane. The region we're interested in is the one enclosed between the two curves, bounded on the left by the line x = 1 and on the right by the line x = 4. This region's area can be found using definite integration, a powerful tool in calculus for calculating areas under curves and between curves.

Key Concepts

Before we proceed, let's review some key concepts:

  • Definite Integral: The definite integral of a function f(x) from a to b, denoted as ∫ab f(x) dx, represents the signed area between the curve of f(x) and the x-axis between the vertical lines x = a and x = b. Areas above the x-axis are positive, and areas below are negative.
  • Area Between Curves: The area between two curves f(x) and g(x) over the interval [a, b] is given by ∫ab |f(x) - g(x)| dx. The absolute value ensures that we're always integrating a positive difference, as area is a positive quantity. To eliminate the absolute value, we need to determine which function is greater over the interval.
  • Natural Logarithm: The natural logarithm, denoted as ln(x), is the logarithm to the base e, where e is an irrational number approximately equal to 2.71828. The natural logarithm function is the inverse of the exponential function e^x.

Setting Up the Integral

The first crucial step in finding the area between the curves is to set up the definite integral correctly. This involves determining the interval of integration and the integrand, which represents the height of the region at each x-value. To do this effectively, we need to identify which function, f(x) or g(x), is the upper bound and which is the lower bound within the interval [1, 4]. Understanding the relationship between the two functions is pivotal for constructing the correct integral expression.

Determining the Upper and Lower Functions

To find out which function is greater, we can analyze the behavior of f(x) = ln(x) + 1 and g(x) = -4x/e + 6 within the interval [1, 4]. We can evaluate both functions at specific points within the interval or use calculus to analyze their derivatives. However, a simple way to compare them is to evaluate them at the endpoints of the interval, x = 1 and x = 4, and at a point in between, say x = e (since the natural logarithm is easy to evaluate at e).

  • At x = 1:
    • f(1) = ln(1) + 1 = 0 + 1 = 1
    • g(1) = -4(1)/e + 6 ≈ -1.47 + 6 = 4.53
  • At x = e:
    • f(e) = ln(e) + 1 = 1 + 1 = 2
    • g(e) = -4(e)/e + 6 = -4 + 6 = 2
  • At x = 4:
    • f(4) = ln(4) + 1 ≈ 1.39 + 1 = 2.39
    • g(4) = -4(4)/e + 6 ≈ -5.89 + 6 = 0.11

From these evaluations, we can see that g(x) is greater than f(x) at x = 1, they are equal at x = e, and f(x) becomes greater than g(x) somewhere between x = e and x = 4. This means the two curves intersect within the interval [1, 4]. To find the exact point of intersection, we set f(x) = g(x) and solve for x:

  • ln(x) + 1 = -4x/e + 6

This equation is transcendental and cannot be solved algebraically. However, we already know that they intersect at x = e. Therefore, g(x) is the upper function from x = 1 to x = e, and f(x) is the upper function from x = e to x = 4. This means we need to split the integral into two parts.

Setting Up the Definite Integrals

Based on our analysis, the area A can be calculated by summing two definite integrals:

  • A = ∫1e [g(x) - f(x)] dx + ∫e4 [f(x) - g(x)] dx

Substituting the functions f(x) and g(x), we get:

  • A = ∫1e [(-4x/e + 6) - (ln(x) + 1)] dx + ∫e4 [(ln(x) + 1) - (-4x/e + 6)] dx
  • A = ∫1e (-4x/e + 5 - ln(x)) dx + ∫e4 (ln(x) + 4x/e - 5) dx

This setup is crucial. We have correctly identified the intervals and the integrand for each interval. The next step is to evaluate these definite integrals.

Evaluating the Integrals

Now that we have the definite integrals set up, we need to evaluate them. This involves finding the antiderivatives of the integrands and applying the Fundamental Theorem of Calculus. The integrals we need to evaluate are:

  • ∫1e (-4x/e + 5 - ln(x)) dx
  • ∫e4 (ln(x) + 4x/e - 5) dx

Evaluating the First Integral

Let's evaluate the first integral, ∫1e (-4x/e + 5 - ln(x)) dx. We can break this into three separate integrals:

  • ∫1e (-4x/e) dx + ∫1e 5 dx - ∫1e ln(x) dx

The first two integrals are straightforward:

  • ∫1e (-4x/e) dx = (-2x^2/e) |1e = (-2e^2/e) - (-2(1)^2/e) = -2e + 2/e
  • ∫1e 5 dx = 5x |1e = 5e - 5

The third integral, ∫1e ln(x) dx, requires integration by parts. Let u = ln(x) and dv = dx. Then du = (1/x) dx and v = x. Using integration by parts, ∫u dv = uv - ∫v du:

  • ∫1e ln(x) dx = [x ln(x)]1e - ∫1e x(1/x) dx = [x ln(x)]1e - ∫1e 1 dx
    •             = [e ln(e) - 1 ln(1)] - [x]1e = [e(1) - 1(0)] - [e - 1] = e - e + 1 = 1*

Putting it all together:

  • ∫1e (-4x/e + 5 - ln(x)) dx = (-2e + 2/e) + (5e - 5) - 1 = 3e + 2/e - 6

Evaluating the Second Integral

Now let's evaluate the second integral, ∫e4 (ln(x) + 4x/e - 5) dx. Again, we break this into three integrals:

  • ∫e4 ln(x) dx + ∫e4 (4x/e) dx - ∫e4 5 dx

We already know ∫ ln(x) dx from the previous calculation. Using integration by parts:

  • ∫e4 ln(x) dx = [x ln(x)]e4 - ∫e4 1 dx = [x ln(x) - x]e4
    •            = [4 ln(4) - 4] - [e ln(e) - e] = 4 ln(4) - 4 - e + e = 4 ln(4) - 4*

The second integral is:

  • ∫e4 (4x/e) dx = (2x^2/e) |e4 = (2(4)^2/e) - (2e^2/e) = 32/e - 2e

The third integral is:

  • ∫e4 5 dx = 5x |e4 = 5(4) - 5e = 20 - 5e

Putting it all together:

  • ∫e4 (ln(x) + 4x/e - 5) dx = (4 ln(4) - 4) + (32/e - 2e) - (20 - 5e) = 4 ln(4) - 24 + 32/e + 3e

Calculating the Total Area

Finally, we add the results of the two integrals to find the total area A:

  • A = (3e + 2/e - 6) + (4 ln(4) - 24 + 32/e + 3e)
  • A = 6e + 34/e - 30 + 4 ln(4)

Therefore, the exact area A of the region bounded between the curve f(x) = ln(x) + 1 and the line g(x) = -4x/e + 6 over the interval [1, 4] is 6e + 34/e - 30 + 4 ln(4) square units.

Final Answer

A = 6e + 34/e - 30 + 4 ln(4) units^2

Summary

In this article, we have meticulously calculated the area of the region bounded by the functions f(x) = ln(x) + 1 and g(x) = -4x/e + 6 over the interval [1, 4]. The process involved setting up definite integrals by determining the upper and lower functions, evaluating these integrals using techniques such as integration by parts, and finally, summing the results to obtain the total area. The exact answer, A = 6e + 34/e - 30 + 4 ln(4) square units, demonstrates the power and precision of calculus in solving geometric problems.

This comprehensive exploration should provide a solid understanding of how to find the area between curves, a skill crucial for success in calculus and its applications.