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The substitution method is a powerful technique used in calculus to simplify complex integrals. It involves changing variables to make an integral more manageable. When you choose a substitution, you select a part of the integrand (the expression under the integral sign) that, when replaced, turns the integral into a simpler form.

For example, in the exercise, the substitution chosen was \( u = \sqrt{x} \). This substitution transforms the complicated integrand into something easier to integrate by using \( u \) instead of \( x \). To use the substitution method:

• Identify a substitution, i.e., a new variable \( u \) that simplifies the integrand.
• Differentiate this new variable with respect to the original variable, i.e., find \( du \).
• Express the original differential in terms of \( du \) and substitute in the integral.
• Integrate with respect to \( u \).
• Finally, substitute back to the original variable, if needed, to get the result in its original terms.

Antiderivatives are essentially the reverse of derivatives. Finding an antiderivative involves determining a function whose derivative is the given function. When computing indefinite integrals, you're essentially finding the antiderivative of the integrand.

In the solution of the given exercise, after the substitution, we obtained \( 2 \int e^{u} \, du \). The function \( e^{u} \) has a known antiderivative, \( e^{u} \), because the derivative of \( e^{u} \) with respect to \( u \) is still \( e^{u} \). Thus, the antiderivative of \( 2e^{u} \) is \( 2e^{u} + C \), where \( C \) is the constant of integration.

Remember that whenever you find an indefinite integral, it includes this constant of integration \( C \) because integration can reverse differentiate multiple functions which differ by only a constant.

Differential calculus focuses on the concept of the derivative, which measures how a function changes as its input changes. This concept is crucial when working with substitutions, as it requires finding derivatives to compute \( du \).

In the context of the given solution, differential calculus is used to find \( du \) from \( u = \sqrt{x} \). Differentiating both sides with respect to \( x \), we have \( \frac{d u}{d x} = \frac{1}{2\sqrt{x}} \). This allows us to express \( du \) as \( \frac{1}{2\sqrt{x}} dx \), which is critical to successfully substituting and eventually integrating the function.

Differential calculus also helps in understanding the behavior of functions through their derivatives, offering insights into the rates at which quantities change, essential to understanding and solving various real-world problems.