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Magazine Chapter 7: Problem 53
Use substitution to evaluate the definite integrals. $$ \int_{5}^{9} \frac{x}{x-3} d x $$
Short Answer
Expert verified
The integral evaluates to \( 4 + 3\ln 3 \).
Step by step solution
01
Choose a Substitution
To evaluate the integral \( \int_{5}^{9} \frac{x}{x-3} \, dx \) using substitution, we need to find a substitution that simplifies the integrand. A good choice is to let \( u = x - 3 \). This means \( du = dx \).
02
Transform the Limits of Integration
The original integral has limits \( x = 5 \) and \( x = 9 \). We need to substitute these into the equation \( u = x - 3 \) to find the new limits in terms of \( u \). \( u(5) = 5 - 3 = 2 \) and \( u(9) = 9 - 3 = 6 \). So the new limits are from \( u = 2 \) to \( u = 6 \).
03
Rewrite the Integral in Terms of \( u \)
Substitute \( u = x - 3 \) and \( x = u + 3 \) into the integral. This gives: \[\int_{2}^{6} \frac{u+3}{u} \, du = \int_{2}^{6} \left( 1 + \frac{3}{u} \right) \, du\].
04
Integrate with Respect to \( u \)
Now, integrate the expression \( \int_{2}^{6} \left( 1 + \frac{3}{u} \right) \, du \). This can be split into two separate integrals:\[\int_{2}^{6} 1 \, du + \int_{2}^{6} \frac{3}{u} \, du\]. The solutions to these are:\[\text{First integral: } \int_{2}^{6} 1 \, du = [u]_{2}^{6} = 6 - 2 = 4\]\[\text{Second integral: } \int_{2}^{6} \frac{3}{u} \, du = 3 \ln|u| \Big|_{2}^{6} = 3(\ln 6 - \ln 2) = 3 \ln\left(\frac{6}{2}\right) = 3 \ln 3\].
05
Solve and Combine the Integrals
Combine the results from Step 4 to evaluate the original integral:\[4 + 3\ln 3\]. This is the final answer to the integral.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
When we talk about definite integrals, we are looking at a way to find the area under a curve between two specific points on the x-axis. The definite integral is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. This process gives us a numerical value representing this area.
- **Definite means exact area**: You are finding the exact area between the function and the x-axis.
- **Limits of integration**: These limits \( a \) and \( b \) help define the start and end points of the interval you鈥檙e considering.
U-Substitution
U-substitution is a technique used to simplify the process of integrating certain functions. It can be thought of as a method for undoing the chain rule of differentiation.
- **Choosing "u" carefully**: The aim here is to pick a "u" that when substituted will simplify the integral significantly. For instance, observe the part that complicates the integrand and see if setting it as \( u \) will simplify the function.
- **Derivatives and differentials**: Once \( u \) is decided (e.g., \( u = x - 3 \)), the next step is to find \( du \) (i.e., the differential \( du = dx \)) to fully make the substitution work.
Transforming Limits of Integration
Transforming the limits is a crucial part of the substitution method when dealing with definite integrals. After making the substitution, the original x-limits need to be converted into u-limits to maintain consistency.
- **Old limits to new limits**: If \( u = x - 3 \), then every x-limit needs the same transformation. For example, if \( x = 5 \), then substitute to find \( u = 5 - 3 = 2 \).
- **Perfect continuity**: This correct transformation results in updated limits where, in our exercise, \( x = 5 \) turns into \( u = 2 \), and \( x = 9 \) turns into \( u = 6 \).
Natural Logarithm Properties
Natural logarithms play an essential role in many integration problems. When integrating functions that involve fractions, such as \( \frac{1}{u} \), you often end up with a natural logarithm as part of the solution.
- **Natural logarithm**: The result of integrating \( \frac{1}{u} \) is \( \ln|u| \). It's important to remember the absolute value in logarithmic functions to ensure the natural log remains valid for negative values as well.
- **Simplifying with logarithms**: Often, properties of logarithms like \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \) can simplify complex results. In our exercise, this property helped in achieving the final answer by simplifying \( 3(\ln 6 - \ln 2) \) to \( 3 \ln 3 \).
