Problem 42 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int \frac{e^{-\sqrt{r}}}{\sqrt{r}} d r\end{equation}

Step-by-Step Solution

Answer

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

Step 1: Introduction to Substitution Method

To evaluate the integral \( \int \frac{e^{-\sqrt{r}}}{\sqrt{r}} \, dr \), we identify that a substitution method can simplify the problem. We aim to simplify the expression using a suitable substitution.

Step 2: Choose a Suitable Substitution

Consider the substitution \( u = \sqrt{r} \), which implies \( du = \frac{1}{2\sqrt{r}} \, dr \). Thus, \( dr = 2u \, du \). This substitution will help simplify both the exponential and the variable \( \sqrt{r} \) in the denominator.

Step 3: Rewrite the Integral with the Substitution

With the substitution \( u = \sqrt{r} \), the integral becomes \( \int e^{-u} \cdot \frac{1}{u} \cdot 2u \, du = 2 \int e^{-u} \, du \). The expression simplifies significantly, allowing us to focus on integrating the exponential function.

Step 4: Integrate the Simplified Expression

Integrate \( 2 \int e^{-u} \, du \) which equals \( -2e^{-u} + C \), where \( C \) is the constant of integration. This result comes from the standard integral of an exponential function.

Step 5: Substitute Back to Original Variable

Recall the substitution \( u = \sqrt{r} \). Replace \( u \) in the integrable function to get the final answer. The final expression in terms of \( r \) is \( -2e^{-\sqrt{r}} + C \).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the concept of integration. Integration allows us to determine the total accumulation of quantities. Whether finding areas, volumes, or solving differential equations, it plays a crucial role. In this exercise, integration helps find the antiderivative of the given function. When approaching integration problems, there are different techniques to consider:

• Substitution Method: Helpful when a function can be transformed into a simpler form.
• Integration by Parts: Useful for functions that are products of two or more functions.
• Partial Fractions: Beneficial when working with rational expressions.

For this particular integral, we utilize substitution to simplify the process. By identifying a part of the integrand to substitute, we transform a complex integral into an easier one, which leads to efficient problem-solving.

Exponential Function

The exponential function is a mathematical function characterized by the expression \( e^x \), where \( e \) is a constant approximately equal to 2.71828. It has special properties making it crucial in calculus, particularly for its derivative and integral properties.Unique Properties of the Exponential Function:

• Derivative: The derivative of \( e^x \) is \( e^x \) itself, which implies its growth rate is proportional to its value.
• Integral: The integral of \( e^x \) is also \( e^x \), showing similar self-similar properties during integration.

In our integral problem, we encounter \( e^{-u} \) as part of the transformed integral. The integral of an exponential function, like \( e^{-u} \), equals \(-e^{-u} + C\). Hence, understanding these fundamental properties aids in efficiently evaluating exponential integrals.

Variable Substitution

Variable substitution is a powerful tool in calculus used to tackle complex integrals. By substituting one variable for another, we can simplify the integrand, making the integral easier to evaluate.

Why Use Substitution?

When an integral contains a composite function or a product of functions, substitution can transform it into a form that is more straightforward to work with. For example, in our problem:- We choose the substitution \( u = \sqrt{r} \), aiming to simplify both the exponential term and the factor in the denominator.- Compute \( du \) to express \( dr \) in terms of \( du \). Here, \( du = \frac{1}{2\sqrt{r}} \, dr \) leads to \( dr = 2u \, du \).

Rewriting the Integral

Using this substitution, the integral \( \int \frac{e^{-\sqrt{r}}}{\sqrt{r}} \, dr \) turns into a simpler form, \( 2 \int e^{-u} \, du \). This transformation removes the square root and exponential complexity, allowing us to integrate easily.To conclude, variable substitution is vital to simplifying and solving integrals that initially appear challenging.