Problem 39 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int_{\ln 4}^{\ln 9} e^{x / 2} d x\end{equation}

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Identify the Integral

We need to evaluate the definite integral \( \int_{\ln 4}^{\ln 9} e^{x/2} dx \). This involves finding the antiderivative of the integrand \( e^{x/2} \).

Step 2: Find an Antiderivative

To find the antiderivative of \( e^{x/2} \), use the substitution method. Let \( u = \frac{x}{2} \), then \( du = \frac{1}{2} dx \) which implies \( dx = 2 du \). Substitute into the integral: \( \int e^{u} \cdot 2 \, du = 2 \int e^{u} \, du \).

Step 3: Integrate the Antiderivative

The integral \( \int e^{u} \, du \) is simply \( e^{u} + C \). Therefore, the antiderivative of \( e^{x/2} \) is \( 2e^{x/2} + C \).

Step 4: Evaluate the Definite Integral

We now evaluate the definite integral \( \int_{\ln 4}^{\ln 9} e^{x/2} dx = \left[ 2e^{x/2} \right]_{\ln 4}^{\ln 9} \). Substitute \( x = \ln 9 \) and \( x = \ln 4 \) into the antiderivative: \( 2e^{(\ln 9)/2} - 2e^{(\ln 4)/2} \).

Step 5: Simplify the Expression

Simplify the terms: \( e^{(\ln 9)/2} = \sqrt{9} = 3 \) and \( e^{(\ln 4)/2} = \sqrt{4} = 2 \). Substitute back: \( 2 \times 3 - 2 \times 2 = 6 - 4 = 2 \).

Key Concepts

Antiderivatives

Antiderivatives often go by another name: indefinite integrals. They are the inverse process of differentiation. Whenever you differentiate a function, you find its rate of change.
This is like reversing that process. If you know the rate of change, the antiderivative gives you the original function.

• The antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \).
• It's important because definite integrals, like in this exercise, rely on finding antiderivatives to evaluate areas under curves.

For our example, the antiderivative of \( e^{x/2} \) is \( 2e^{x/2} \). This is crucial for evaluating the definite integral from \( \ln 4 \) to \( \ln 9 \).

Substitution Method

The substitution method is a powerful technique dealing with more complex integrals by changing variables. It simplifies the integrand, making it easier to integrate. Think of it like changing the perspective to see the solution more clearly.

• Choose a substitution that will simplify the expression, like \( u = \frac{x}{2} \) in our exercise.
• Rewrite \( dx \) according to \( du \), such as \( dx = 2 \, du \).
• Your integral then becomes something like \( \int e^u \, 2 \, du \), which is much simpler.

After solving the integral in terms of \( u \), substitute back the original terms to get your solution in the original variable.

Logarithmic Integration

Logarithmic integration usually takes a special form, but it's essential for definite integration when the bounds involve logarithms. Understanding the properties of logarithms can make solving integrals with log limits much easier.

• Logarithms simplify when used as exponents: \( e^{\ln a} = a \).
• This fact was used to simplify \( e^{(\ln 9)/2} \) to \( \sqrt{9} \), or simply 3, in this exercise.
• Such simplifications make evaluating the definite integral straightforward.

Remember, using the inverse properties between exponential functions and logarithms allows for quick simplifications.

Exponential Functions

Exponential functions, like \( e^x \), are a cornerstone in calculus because of their unique properties. They reappear as their own derivatives and integrals, which is a fascinating trait.

• The derivative and integral of \( e^x \) are both \( e^x \), simplifying many calculations.
• In integrals, sometimes you just need to adjust coefficients to match the function form, e.g., multiplying by 2 in \( 2 \int e^u du \).
• Being aware of these properties allows you to handle exponential terms gracefully in integration.

By manipulating exponential expressions correctly, like transforming \( e^{x/2} \) to \( e^u \), you make them much less intimidating to integrate.