Problem 29 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 21 through 30 , evaluate the indicated definite integral. \(\int_{0}^{e-1}\left(\frac{x}{x+1}\right) d x\)

Step-by-Step Solution

Answer

The definite integral evaluates to \(e - 2\).

Step 1: Set up the Integral

The given integral is \ \ \ \(\int_{0}^{e-1}\left(\frac{x}{x+1}\right) dx\). Begin by identifying the limits of integration and the integrand.

Step 2: Use Substitution

Let \(u = x + 1\). Then \(du = dx\) and \(x = u - 1\). Adjust the limits accordingly: When \(x = 0\), \(u = 1\); and when \(x = e-1\), \(u = e\).

Step 3: Substitute and Simplify

Substitute \(x\) and \(dx\) in the integral: \ \ \(\int_{0}^{e-1} \frac{x}{x+1} dx \rightarrow \int_{1}^{e} \frac{u-1}{u} du\). Split the integral: \ \ \(\int_{1}^{e} \frac{u}{u} du - \int_{1}^{e} \frac{1}{u} du\)

Step 4: Integrate

Evaluate each part separately: \ \ \(\int_{1}^{e} 1 du - \int_{1}^{e} \frac{1}{u} du\) \ \ The first integral: \(\int_{1}^{e} 1 du = [u]_{1}^{e} = e - 1\). \ \ The second integral: \(\int_{1}^{e} \frac{1}{u} du = [\ln|u|]_{1}^{e} = \ln|e| - \ln|1| = 1 - 0 = 1\).

Step 5: Combine Results

Subtract the two results: \ \ \( (e - 1) - 1 = e - 2\)

Key Concepts

integration by substitution

Integration by substitution is a powerful technique to evaluate definite integrals, especially when the integrand involves composite functions. Think of it as a reverse chain rule. You simplify the integrand by substituting a part of it with a new variable.

In the given exercise, we begin by letting \( u = x + 1 \). This substitution is chosen because the derivative of \( u \) (which is \( du = dx \)) will help simplify the integral.

Changing the limits of integration according to the substitution is crucial. When \( x = 0 \), \( u = 0 + 1 = 1 \), and when \( x = e-1 \), \( u = (e-1) + 1 = e \). This changes the integral to use these new limits, making the subsequent steps easier to handle.

Finally, substitute the expression inside the integral and the differential \( dx \) with \( du \). Now, the integral to solve is \( \int_{1}^{e} \frac{u-1}{u} du \), which you can split further for easier evaluation.

evaluating integrals

Once we have simplified our integral using substitution, we move on to the evaluation step.

The new integral \( \int_{1}^{e} \frac{u-1}{u} du \) can be split into two simpler integrals: \( \int_{1}^{e} \frac{u}{u} du - \int_{1}^{e} \frac{1}{u} du \).

This gives us two integrals:

• \( \int_{1}^{e} 1 du \), which is an integral of a constant function
• \( \int_{1}^{e} \frac{1}{u} du \), which is a common logarithmic integral

The first integral evaluates to \( u \) over the given limits. Plugging in the limits, we get \( e - 1 \).

The second integral, involving \( \frac{1}{u} \), evaluates to \( \ln|u| \). Plugging in the limits, we get \( \ln|e| - \ln|1| = 1 - 0 = 1 \).

Combining these results, we finally get: \( (e - 1) - 1 = e - 2 \).

limits of integration

When dealing with definite integrals, the limits of integration play a critical role. They define the range over which we need to evaluate the integral.

After making a substitution, it is important to adjust the limits of integration to correspond with the new variable.

For the given exercise, after substituting \( u = x + 1 \), the limits change as follows:

• When \( x = 0 \), the limit changes to \( u = 1 \)
• When \( x = e-1 \), the limit changes to \( u = e \)

This means that instead of integrating from \( 0 \) to \( e-1 \) in terms of \( x \), we now integrate from \( 1 \) to \( e \) in terms of \( u \).

This adjustment ensures that we are evaluating the integral correctly over the same range, just with a different variable. Remember, incorrect limits can completely change the value of your definite integral, so always pay attention to this step when using the substitution method.