91Ó°ÊÓ

The term "antiderivative" refers to a function whose derivative results in the given function. In other words, if we differentiate an antiderivative, we should obtain the original function back.
For instance, in the equation \ \( \int \left( \frac{1}{x^2} - x^2 - \frac{1}{3} \right) \, dx \), we're looking for a function that when derived gives us \( \frac{1}{x^2} - x^2 - \frac{1}{3} \).
Finding an antiderivative means integrating a function, which involves reversing the differentiation process.
This is why indefinite integrals are sometimes referred to as finding the antiderivative.
The solution includes a constant \( C \), because derivatives of constants are zero, meaning any constant added would still yield the same original function when differentiated.

To successfully integrate a function like \( \int \left( \frac{1}{x^2} - x^2 - \frac{1}{3} \right) \, dx \), we often break it down into simpler components.
This makes the process more manageable since many functions can be integrated using standard rules.

• First, we rewrite terms for easier manipulation: expressing \( \frac{1}{x^2} \) as \( x^{-2} \) using exponent laws.
• Next, we integrate each term separately:
  • The first term \( x^{-2} \) integrates to \(-x^{-1}\).
  • The second term \(-x^2\) integrates to \(-\frac{x^3}{3}\).
  • The third term \(-\frac{1}{3}\) becomes \(-\frac{1}{3}x\) when integrated.

By integrating each, we obtain the general antiderivative:
\(-x^{-1} - \frac{x^3}{3} - \frac{1}{3}x + C\), where \( C \) is the constant of integration.
This approach helps simplify the integration process through the power rule, applicable to polynomials.

Verification of our calculations involving indefinite integrals is a crucial step.
We confirm the antiderivative as correct by differentiating it to see if it gives us our original function.
Consider the antiderivative found: \(-x^{-1} - \frac{x^3}{3} - \frac{1}{3}x + C\).

• Differentiating \(-x^{-1}\) gives \(x^{-2}\), recovering our first term.
• Differentiating \(-\frac{x^3}{3}\) yields \(-x^2\), matching the second term.
• Differentiating \(-\frac{1}{3}x\) gives \(-\frac{1}{3}\), equivalent to the third term.

By these calculations, each step of differentiation confirms it returns us to the original function: \( \frac{1}{x^2} - x^2 - \frac{1}{3} \).
This demonstrates that our integral and thus our antiderivative is accurate and reliable.
This verification procedure reinforces the completeness and correctness of the integration process in calculus.