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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.

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 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.