Problem 24 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{-2}^{2} \frac{e^{x / 2}}{e^{x / 2}+1} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $\int_{-2}^{2} \frac{e^{x / 2}}{e^{x / 2}+1} dx$. Answer: The definite integral evaluates to $2(e - e^{-1} + \ln(e^{-1}+1) - \ln(e+1))$.

Step 1: Rewrite the integrand

To simplify the integrand, we can make a substitution. Let \(u = e^{x/2} + 1\). This means that \(du = (1/2)e^{x/2} dx\). Now we can rewrite the integrand as: $$\frac{e^{x / 2}}{e^{x / 2}+1} = \frac{u-1}{u} = 1 - \frac{1}{u}$$ We also need to change the limits of integration to correspond to the new variable \(u\). When \(x=-2\), \(u=e^{-1} + 1\). When \(x=2\), \(u=e^1 + 1\). So our integral now looks like this: $$\int_{e^{-1}+1}^{e^1+1} (1 - \frac{1}{u})\cdot 2 du$$ Notice that we multiplied by \(2\) because \(du = (1/2)e^{x/2} dx\), so \(2du = e^{x/2} dx\).

Step 2: Integrate the new integrand

Now we can integrate the function with respect to \(u\). We obtain: $$2\int_{e^{-1}+1}^{e^1+1} (1 - \frac{1}{u}) du = 2\left[\int_{e^{-1}+1}^{e^1+1} du - \int_{e^{-1}+1}^{e^1+1} \frac{1}{u} du\right]$$ The first integral is simply the integral of 1 with respect to \(u\), which is \(u\), and the second integral is the natural logarithm, \(\ln|u|\). So, we have: $$2\left[u - \ln|u|\right]_{e^{-1}+1}^{e^1+1}$$

Step 3: Calculate the definite integral

Finally, we will calculate the definite integral using the new expression: $$2\left[u - \ln|u|\right]_{e^{-1}+1}^{e^1+1} = 2\left[\left(e^{1}+1 - \ln(e^{1}+1)\right) - \left(e^{-1}+1 - \ln(e^{-1}+1)\right)\right]$$ Simplify the expression: $$2\left[(e+1-\ln(e+1)) - (e^{-1}+1-\ln(e^{-1}+1))\right]$$

Step 4: Simplify and find the final answer

Now we can simplify the expression and find the final answer: $$2\left[(e+1-\ln(e+1)) - (e^{-1}+1-\ln(e^{-1}+1))\right] = 2(e - e^{-1} + \ln(e^{-1}+1) - \ln(e+1))$$ The final answer to the integral is: $$2(e - e^{-1} + \ln(e^{-1}+1) - \ln(e+1))$$

Key Concepts

Definite Integrals

Definite integrals in calculus are used to find the area under the curve of a function over a specific interval on the x-axis. In our problem, we are asked to find the area of the curve represented by the integral \[\int_{-2}^{2} \frac{e^{x / 2}}{e^{x / 2}+1} d x\].

This function is an exponential function and the integral is evaluated from -2 to 2. The notation \(\int_{a}^{b} f(x) \, dx\) denotes a definite integral from \(a\) to \(b\).

• The endpoints \(a\) and \(b\) represent the limits of integration and define the interval.
• The function \(f(x)\) is called the integrand.

The result of a definite integral is a number, not a function, and it can be thought of as the signed area between the x-axis and the graph of the integrand. In this case, using other calculus methods such as substitution, we solve this integral to find the exact area under our curve from \(x = -2\) to \(x = 2\).

Substitution Method

The substitution method is a crucial technique in integral calculus. It simplifies an integral by making a substitution to turn it into a form that is easier to integrate. This method is also known as "u-substitution."

For our integral, we made the substitution \(u = e^{x/2} + 1\). Why do we choose this substitution?

• This choice simplifies the integrand into a form easier to integrate, i.e., \(1 - \frac{1}{u}\).
• It requires us to adjust the differential, from \(dx\) to \(du\), using the derivative of the substitution. Here, \(du = \frac{1}{2}e^{x/2} dx\).
• The limits of integration also change according to the new variable \(u\):
  
  • When \(x = -2\), \(u = e^{-1} + 1\).
  • When \(x = 2\), \(u = e^{1} + 1\).

This technique makes the integral more straightforward to evaluate and is particularly useful when dealing with complex functions such as exponential functions.

Exponential Functions

Exponential functions are a key part of this integral problem. They are functions of the form \(f(x)= a^x\) where \(a\) is greater than zero and typically \(e\) (the base of natural logarithms) is used for continuous growth models.

In the context of our integral, we have \(e^{x/2}\), which is an expression containing the exponential function base \(e\). Why is understanding exponential functions important here?

• These functions have unique properties including their slopes and rate of growth, making them very important in many areas of mathematics.
• The presence of the exponent requires us to apply specific calculus techniques such as substitution and integration rules that handle this rate of change effectively.
• Knowing how to manipulate and simplify exponential expressions aids in solving complex integrals such as our original problem.

The evaluation and simplification steps of integral calculus are heavily reliant on the characteristics of exponential functions, which help us find precise areas under curves or solve pertinent real-world problems.