91Ó°ÊÓ

Integration is a core concept in calculus that is used to find the original function from its derivative. In mathematical terms, when you integrate a function, you are essentially reversing the process of differentiation. This is why integration is sometimes referred to as the antiderivative.
In the exercise we are examining, we deal with the function's derivative: \( f'(x) = -2x \sin x^2 \). Our goal is to obtain the original function \( f(x) \) by integrating \( f'(x) \). The integral of a function typically involves finding an expression whose derivative would yield the original function given. This process is crucial in solving problems related to areas under curves, accumulation of quantities, and reversing differentiation.
Overall, mastering integration means enhancing your calculus toolset for a wide range of applications.

The substitution method is an essential technique in integration, particularly useful when dealing with complex integrals. Often, directly integrating a function can be overly complicated or impossible, so substitution simplifies the integral by changing variables.
In our problem, we substitute \( u = x^2 \) and consequently, \( du = 2x \, dx \) is our differential. The original integral \( \int -2x \sin(x^2) \, dx \) becomes \( \int -\sin(u) \, du \) after substitution. This new integral is more straightforward to evaluate.

• Substitution works by choosing a part of the integrand (the function being integrated) to replace with a single variable.
• The differential \( du \) replaces \( dx \) based on this substitution.
• After integrating in terms of \( u \), always remember to substitute back in terms of the original variable \( x \).

The method highlights an essential aspect of integration: transforming complex problems into simpler ones where the integral can be directly calculated.

Finding the general solution involves determining the function that encompasses all possible solutions to the given differential equation. In integration, this is represented by the constant of integration, \( C \), since any constant disappears when differentiating.
For our exercise, the general solution to \( \int -\sin(u) \, du \) is \( \cos(u) + C \). After substituting back \( u = x^2 \), we yield the solution \( f(x) = \cos(x^2) + C \).
It's important to keep in mind:

• The constant \( C \) accounts for any vertical translation in the graph of the integral function, ensuring that all possible shifts (solutions) are considered.
• In initial value problems, specific values for\( C \) can be determined to find particular solutions that satisfy certain conditions.
• The general solution is critical for understanding the complete family of solutions a differential equation may have.

The concept of a general solution plays a significant role in differential equations and various fields such as physics and engineering.