Problem 38 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(29-50.\) $$\int \frac{e^{-\sqrt{r}}}{\sqrt{r}} d r$$

Step-by-Step Solution

Answer

The integral evaluates to \( -2e^{-\sqrt{r}} + C \).

Step 1: Substitution Step

Use substitution to simplify the integral. Let \( u = \sqrt{r} \), which implies \( r = u^2 \). Thus, the derivative \( dr = 2u \, du \). Substitute these into the integral to transform it into terms of \( u \).

Step 2: Substitute and Simplify

With the substitution, the integral \( \int \frac{e^{-\sqrt{r}}}{\sqrt{r}} \, dr \) becomes \( \int \frac{e^{-u}}{u} \cdot 2u \, du \). The \( u \) in the denominator and the \( u \) in \( 2u \, du \) cancel each other out, resulting in \( 2 \int e^{-u} \, du \).

Step 3: Integrate the Simplified Expression

Now, integrate the simplified expression \( 2 \int e^{-u} \, du \). The integral of \( e^{-u} \) is \( -e^{-u} \), thus we have \( 2(-e^{-u}) + C \), where \( C \) is the integration constant. This results in \( -2e^{-u} + C \).

Step 4: Back-substitution

Perform the back-substitution to express the result in terms of \( r \). Remember \( u = \sqrt{r} \), so substitute back to get \( -2e^{-\sqrt{r}} + C \).

Key Concepts

Substitution Method

When faced with complex integrals, the substitution method proves to be a powerful tool. It simplifies the process by making a clever change of variables. In the given exercise, we started by letting \( u = \sqrt{r} \). This choice of substitution is strategic:

• The derivative \( dr = 2u \, du \) helps align differentials.
• The expression \( r = u^2 \) allows us to convert the integral fully in terms of new variables.

This method essentially transforms a complex-looking integral into a more manageable form. The initial substitution reduces the given integral into simpler pieces, removing tricky parts and turning them into standard integral forms.
Overall, substitution helps in streamlining the computation and, more importantly, uncovers the underlying simplicity of the integral, which might not be apparent at first glance.

Exponential Function

Exponential functions, particularly those such as \( e^{-x} \), are fundamental in calculus due to their unique properties. They remain unchanged in terms of their derivatives and integrals except for a constant factor. Here, an exponential term \( e^{-u} \) emerges after substitution:

• Exponential functions significantly simplify the integration process due to straightforward derivatives and antiderivatives.
• The integral \( \int e^{-u} \, du \) is easy to compute, resulting in \( -e^{-u} \).

In practical terms, exponential functions enable smooth progress from step to step. They often appear in calculus, embodying scenarios of exponential growth or decay, and understanding them is crucial to mastering integral calculus.
This case emphasizes the simplicity yet power of exponential functions and how they facilitate solving integrals involving exponential terms.

Indefinite Integrals

Indefinite integrals represent a class of problems centered around finding antiderivatives. Unlike definite integrals which have limits, indefinite integrals require determining a function \( F(x) \) that fits under the derivative \( f(x) \). In this solution:

• The result \( -2e^{-\sqrt{r}} + C \) is an indefinite integral, meaning it includes an arbitrary constant \( C \).
• Integration remains a process of reversing differentiation, which is central to analyzing functions in calculus.

Adding \( C \) accounts for the fact that the antiderivative is not unique; infinitely many functions differ by a constant. Calculus students must grasp this concept to appreciate the full breadth of indefinite integrals.
Recognizing the need for the constant of integration and its role is essential for accurate and comprehensive solutions in integral calculus.