Problem 14 - Calculus and its applications

Question

Evaluate. (Be sure to check by differentiating!) $$ \int e^{x / 2} d x $$

Step-by-Step Solution

Answer

The integral is \( 2e^{x/2} + C \).

Step 1: Identify the Integral Form

The integral given is \( \int e^{x/2} \, dx \). This function looks like an exponential function of the form \( e^{u} \), where \( u = \frac{x}{2} \). Our task is to express this function in a standard integrable form.

Step 2: Substitute to Simplify Integration

Let \( u = \frac{x}{2} \). Then the derivative \( \frac{du}{dx} = \frac{1}{2} \), or equivalently, \( du = \frac{1}{2} dx \), which implies \( dx = 2 du \). Substitute these into the integral to get \( \int e^{u} \cdot 2 \, du \).

Step 3: Integrate the Simplified Form

Now integrate using the standard integral formula for exponentials: \( \int e^{u} \, du = e^{u} \). Thus, our integral becomes \( 2 \int e^{u} \, du = 2e^{u} + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back to Original Variables

Now replace \( u \) with \( \frac{x}{2} \) to express the antiderivative in terms of \( x \): \( 2e^{u} + C = 2e^{x/2} + C \).

Step 5: Check the Solution by Differentiating

Differentiate the expression \( 2e^{x/2} + C \) with respect to \( x \): Using the chain rule, \( \frac{d}{dx}[2e^{x/2}] = 2 \cdot e^{x/2} \cdot \frac{1}{2} = e^{x/2} \). This is the original integrand, confirming that the solution is correct.

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in calculus and mathematics broadly. They are expressions where a base, typically the constant \( e \), is raised to the power of a variable. This base, \( e \), is an irrational number approximately equal to 2.71828. Exponential functions have unique properties that make them particularly interesting:

• They grow at rates proportional to their current value, which means if you plot an exponential function, it will steadily increase or decrease, never reaching a zero slope.
• One of their defining characteristics is that the derivative of an exponential function is proportional to the function itself.

In the exercise, we have the integral \( \int e^{x/2} \, dx \), taking the base \( e \) to the exponent \( x/2 \). By understanding these characteristics of exponential functions, it becomes easier to apply integration techniques.

Substitution Method

The substitution method is a powerful tool for simplifying integration, especially when dealing with complex functions. It involves transforming part of the original integral into a new variable to make the function easier to integrate.
Here's how it works:

• Identify a part in the integrand that can be substituted, usually something that makes the integral less complicated.
• Set this part equal to a new variable, say \( u = \frac{x}{2} \).
• Determine the differential of this new variable. In the case of our example, since \( u = \frac{x}{2} \), then \( du = \frac{1}{2} \, dx \), or \( dx = 2 \, du \).

Replace these into the integral, changing the variable and factors. Now, instead of integrating \( e^{x/2} \, dx \), we reframe it as \( 2 \int e^u \, du \), which is much simpler. This simplification is key, as it turns complex tasks into manageable steps.

Antiderivative Verification

Antiderivative verification is an essential step after finding an integral to ensure that the solution is correct. This checks if the antiderivative, when differentiated, returns to the original function you started with.
Here's how you verify:

• Once you have found the antiderivative (in our case, \( 2e^{x/2} + C \)), differentiate it with respect to the original variable \( x \).
• Apply differentiation rules, like the chain rule if necessary. Here, differentiating \( 2e^{x/2} \) involves multiplying by the derivative of the exponent, which is \( \frac{1}{2} \).
• So the derivative becomes \( 2 \times e^{x/2} \times \frac{1}{2} = e^{x/2} \).

This should match the original integrand of the exercise. Therefore, verifying the antiderivative helps to confirm that your initial solution is indeed correct and fully validated.