Problem 29 - Applied Calculus

Question

Use an integral to find the specified area. Under \(y=5 \ln (2 x)\) and above \(y=3\) for \(3 \leq x \leq 5\).

Step-by-Step Solution

Answer

The area is approximately 2.554 square units.

Step 1: Set Up Integral for Area

The area under the curve and above the line is given by the integral of the difference between the two functions. We start by setting up the integral from the lower limit 3 to the upper limit 5: \[ A = \int_{3}^{5} (5 \ln(2x) - 3) \, dx \]

Step 2: Integrate 5 ln(2x)

We need to evaluate the integral of the function \(5 \ln(2x)\). To do this, use the substitution \(u = 2x\), \(du = 2 \, dx\), so \(dx = \frac{du}{2}\). This transforms the integral: \[ \int 5 \ln(2x) \, dx = \frac{5}{2} \int \ln(u) \, du \] Integrate \(\ln(u)\) using integration by parts with \(v = \ln(u), dv = \frac{1}{u} du\) and \(w' = 1\), so \(w = u\): \[ \int \ln(u) \, du = u(\ln(u) - 1) \] Evaluating the definite integral, transform back: \[ \frac{5}{2} \int \ln(u) \, du = \frac{5}{2} [u(\ln(u) - 1)] \ = \frac{5}{2} [2x(\ln(2x) - 1)] = 5x \ln(2x) - 5x \]

Step 3: Integrate 3

The integral of 3 over \([3,5]\) is straightforward. The integral is: \[ \int_3^5 3 \, dx = 3x \] Evaluating this from 3 to 5 gives:\[ 3[5 - 3] = 6 \]

Step 4: Evaluate the Definite Integral

Substitute the results from steps 2 and 3 into the original integral \[ A = \left[5x \ln(2x) - 5x\right]_{3}^{5} - \left[3x\right]_{3}^{5} \] Calculate the value of each part:\[ [5(5) \ln(10) - 25] - [5(3) \ln(6) - 15] = (25 \ln(10) - 25) - (15 \ln(6) - 15) \]\[ 5 ( \ln(10) - \ln(6) ) = 5 \ln(\frac{10}{6}) = 5 \ln(\frac{5}{3}) \] Thus, \(A = 5 \ln(\frac{5}{3})\).

Step 5: Final Area Calculation

Calculate the final value numerically: \[ A \approx 5 \ln\left(\frac{5}{3}\right) \approx 5 \times 0.5108 \approx 2.554 \] Thus, the area is approximately 2.554 square units.

Key Concepts

Definite Integrals

Definite integrals play a crucial role in calculus, allowing us to find the exact area under a curve between two points on the x-axis. In the context of our exercise, we use a definite integral to find the area between the curve described by the function \( y = 5 \ln (2x) \) and the line \( y = 3 \), from \( x = 3 \) to \( x = 5 \).

When setting up a definite integral, you involve limits of integration that define the start and end points along the x-axis. In this case, our limits are from 3 to 5. The definite integral \[ \int_{3}^{5} (5 \ln(2x) - 3) \, dx \] easily finds the area by integrating the difference between the curve and the horizontal line. This involves subtracting the function representing the line from the function of the curve.

• This subtraction is crucial because we need the net area confined by both the functions, not the whole area under the curve.
• When evaluated, the definite integral gives a numerical value representing the area enclosed.

The art of computing definite integrals lies in careful setup and calculation, as seen in the steps of integration by parts and substitution techniques.

Area under a Curve

Understanding the "area under a curve" is fundamental in calculus, especially in definite integrals. The concept essentially means calculating the space between a function graph and a horizontal line, often the x-axis. For our specific problem, the area lies between \( y = 5 \ln (2x) \) and the straight line \( y = 3 \).

Calculating this area involves using the integral to sum up infinitesimally small strips of width \( dx \) and height given by the difference between the two functions over the interval \([3,5]\).

• By focusing on the distinct strip-by-strip approach of integrals, students can visualize how calculus aggregates these to form a whole area.
• Knowing the integral limits

devises what part of the curve contributes to the area.

Ultimately, the integral tells us how much total area those tiny strips encompass between the curve and the specified line. This specific method beautifully showcases why integration is so powerful in areas such as physics and engineering.

Integration by Parts

Integration by parts is an invaluable technique when dealing with integrals that are products of functions. It's a bit like the reverse of the product rule for differentiation. This method becomes essential for evaluating integrals that are not straightforward, like \( \int \ln(u) \, du \) in our exercise.

The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]This equation allows us to convert a difficult integral into simpler parts. For example:

• We select \( v = \ln(u) \) and \( du = \frac{1}{u} \, du \)
• This helps in reducing the complexity of finding the antiderivative of logarithmic functions.

During this exercise, here's how integration by parts unfolds:

• Using \( u = 2x \) as a substitution simplifies \( dx = \frac{du}{2} \).
• This setup transforms our integral to one infiltrated by a simpler form: \( \int \ln(u) \, du \).

By following through this methodical approach, we disentangle the initially intricate integral to one manageable through basic integration techniques.