Problem 26 - Applied Calculus

Question

Use an integral to find the specified area. Between \(y=\cos x+7\) and \(y=\ln (x-3), 5 \leq x \leq 7.\)

Step-by-Step Solution

Answer

The area between the curves is the result of the evaluated integral, found numerically.

Step 1: Identify the Intersection Points

To find the area between the functions, you need to know where they intersect. Set the functions equal to each other: \[ \cos(x) + 7 = \ln(x - 3) \]Solve numerically or graphically since the analytical solution might be complex, within the interval 5 to 7. Generally use a graphing calculator or computer algebra system to find the intersection points.

Step 2: Set Up the Integral

The functions intersect over the interval where both are defined. So, we'll integrate from where the functions intersect within the interval. Because \( y = \cos(x) + 7 \)is the upper curve and\( y = \ln(x-3) \)is the lower curve in this interval. Set up the integral:\[ A = \int_{5}^{7} [(\cos(x) + 7) - \ln(x-3)] \, dx \]

Step 3: Calculate the Integral

Now, calculate the definite integral:\[ A = \int_{5}^{7} \cos(x) \, dx + \int_{5}^{7} 7 \, dx - \int_{5}^{7} \ln(x-3) \, dx \]Evaluate each part separately:- For \( \int \cos(x) \, dx = \sin(x) \), - For \( \int 7 \, dx = 7x \), - For \( \int \ln(x-3) \, dx \), use integration by parts.

Step 4: Evaluate and Combine Results

Evaluate each integral:- \( \left[ \sin(x) \right]_{5}^{7} = \sin(7) - \sin(5) \)- \( \left[ 7x \right]_{5}^{7} = 7 \times 7 - 7 \times 5 \)- For \( \int \ln(x-3) \, dx \), use integration by parts where \( u = \ln(x-3) \) and \( dv = dx \): \[ \int u \, dv = uv - \int v \, du \] Evaluate this part and then compute the difference.

Step 5: Compute the Area Result

Sum up the evaluated results: - Combine the results from Step 4 to get the area between the curves over the given interval. Calculations should yield a numerical value which is the area.

Key Concepts

Definite Integral

A definite integral is a powerful tool in calculus that calculates the net area under a curve from one point to another along the x-axis. This area can represent various physical quantities such as distance, mass, or, in our exercise, the space between two curves. When you see an integral symbol with limits at the top and bottom, like \[ \int_{a}^{b} f(x) \, dx \]this is indicating a definite integral. Here, \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated.

### Evaluating a Definite IntegralTo evaluate a definite integral, perform these steps:

• Find the antiderivative (indefinite integral) of the function \( f(x) \).
• Evaluate this antiderivative at the upper limit \( b \) and at the lower limit \( a \).
• Subtract the value of the antiderivative at \( a \) from its value at \( b \).

In the given exercise, the definite integral represents the area enclosed by the two functions, \( y=\cos(x)+7 \) and \( y=\ln(x-3) \). Thus, we established the integral from 5 to 7 and evaluated it to find this area.

Area Between Curves

Finding the area between two curves is a common application of definite integrals in calculus. This involves integrating the difference between the two functions over the interval where they intersect.
### Understanding the ConceptThe basic formula to find the area \( A \) between two curves given by \( y=f(x) \) and \( y=g(x) \) over an interval \([a,b] \) is:\[ A = \int_{a}^{b} [f(x) - g(x)] \, dx \]In this calculation:

• \( f(x) \) is the function on top (upper curve).
• \( g(x) \) is the function on the bottom (lower curve).

The area is determined by subtracting the lower curve from the upper curve and integrating over the desired interval.

In our exercise, \( \cos(x) + 7 \) serves as the upper curve and \( \ln(x-3) \) as the lower curve. We integrate from \( x = 5 \) to \( x = 7 \) to find the area between them.

Integration by Parts

Integration by Parts is a technique derived from the product rule for differentiation. It is quite useful for integrating products of functions, particularly when one function is easier to integrate than the other. The formula is given by:\[ \int u \, dv = uv - \int v \, du \]where \( u \) and \( dv \) are parts of the original integrand.
### Applying Integration by PartsTo use this method effectively:

• Choose \( u \) to be the part of the integrand that becomes simpler when differentiated.
• Choose \( dv \) such that \( v \) (found by integrating \( dv \)) is not more complex than \( dv \).

In the exercise, we apply this method to integrate \( \ln(x-3) \), where we let \( u = \ln(x-3) \) and \( dv = dx \). This way, differentiating \( u \) gives a simple expression, making the integration process more manageable. Using this technique, we integrate each part, calculate, and combine them to find the total area.