91Ó°ÊÓ

The substitution method is a powerful technique in calculus, particularly useful when dealing with complex integrands, like exponential functions involving square roots. In the exercise given, we simplify the integral \( \int e^{\sqrt{3s + 9}} \, ds \) by choosing a suitable substitution. The idea is to replace a part of the integrand to make the integral more tractable.
Let's choose \( u = \sqrt{3s + 9} \). This substitution transforms the problem by turning the complex expression under the integrand into a simpler variable, \( u \).

• By expressing \( 3s + 9 \) as \( u^2 \), this substitution simplifies the expression within the square root.
• Differentiating both sides with respect to \( s \), and rearranging it allows us to express \( ds \) in terms of \( du \).
• This results in \( ds = \frac{2u}{3} \, du \), which is key to replacing \( ds \) in the integral.

Substitution transforms our original integral into one involving \( u \): \( \frac{2}{3} \int u e^u \, du \). This simplification makes the integral easier to handle with additional calculus methods, like integration by parts.

Understanding differentiation in the context of substitution is essential to calculus problems. Differentiating the substituted variable \( u = \sqrt{3s + 9} \) with respect to \( s \) is a crucial step.
The differentiation gives us a relationship between \( du \) and \( ds \):

• First, express \( u \) squared: \( u^2 = 3s + 9 \).
• By differentiating both sides with respect to \( s \), we get \( 2u \frac{du}{ds} = 3 \).
• Solving for \( ds \), we find \( ds = \frac{2u}{3} \, du \).

This relationship allows us to convert the integration variable from \( s \) to \( u \), making the integral simpler and more manageable. Remember, differentiating correctly and substituting back are keys to ensuring the integrity of the integral transformation and eventual solution.

The antiderivative is the result of integrating a function. It's the function we obtain when reversing the process of differentiation. In this problem, obtaining the antiderivative involves several steps:
Once the integral has been transformed by substitution, we find the antiderivative of the resulting expression.
The transformed integral is \( \frac{2}{3} \int u e^u \, du \).
Through integration by parts, we find the antiderivative:

• Let's choose \( v = u \) and \( dw = e^u \, du \).
• Using the integration by parts formula, \( \int v \, dw = vw - \int w \, dv \), we calculate \( \int u e^u \, du = u e^u - \int e^u \, du \).
• Then, solve \( \int e^u \, du \), which is \( e^u \).

Hence, the antiderivative of the initial integral becomes \( \frac{2}{3} (u e^u - e^u) + C \). Finally, substitute \( u = \sqrt{3s + 9} \) back to express the antiderivative in terms of \( s \). Resulting in: \( \frac{2}{3} e^{\sqrt{3s + 9}} (\sqrt{3s + 9} - 1) + C \).

The exercise combines multiple calculus integration techniques to solve a complex integral. These techniques each harness specific rules and transformation processes:
1. **Substitution Method**: A key technique that turns difficult integrals into more manageable forms by changing variables. It simplifies the expression significantly before applying other methods.
2. **Integration by Parts**: This technique is particularly useful when the integrand is a product of functions, like \( u e^u \). By choosing parts of the integrand to integrate and differentiate, integration by parts breaks the integral into more straightforward segments.
3. **Evaluation of Simple Integrals**: After transforming and simplifying, you often end up with integrals of common functions, like \( \int e^u \, du \), which offer straightforward solutions.

• Integration by parts and substitution are complementary and can often be used in tandem to simplify complex calculus problems.
• Knowing each technique allows you to recognize when to apply them effectively, depending on the structure of the integrand.

Each step in the integration process helps to progressively simplify and solve complex integrals, ultimately revealing the underlying functions hidden behind layers of mathematical expression.