Problem 23 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int\left(\frac{1}{x^{2}}-x^{2}-\frac{1}{3}\right) d x $$

Step-by-Step Solution

Answer

The most general antiderivative is \(-x^{-1} - \frac{x^3}{3} - \frac{1}{3}x + C\).

Step 1: Break Down the Integral

The given integral is: \[ \int \left( \frac{1}{x^2} - x^2 - \frac{1}{3} \right) \, dx \]We can rewrite the terms for easier integration: \[ \int \left( x^{-2} - x^2 - \frac{1}{3} \right) \, dx \]This allows us to consider each term separately.

Step 2: Integrate Each Term Separately

**Integrate the first term**: The integral of \( x^{-2} \) is: \[ \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -x^{-1} \]**Integrate the second term**: The integral of \( -x^2 \) is: \[ \int -x^2 \, dx = -\frac{x^3}{3} \]**Integrate the third term**: The integral of \( -\frac{1}{3} \) is: \[ \int -\frac{1}{3} \, dx = -\frac{1}{3}x \]

Step 3: Combine the Results

Combine the results from each term to form the general antiderivative:\[ -x^{-1} - \frac{x^3}{3} - \frac{1}{3}x + C \]where \( C \) is the constant of integration.

Step 4: Verify by Differentiation

Differentiate the obtained antiderivative to check if we get back to the original function:- Differentiate \( -x^{-1} \) to get \( x^{-2} \).- Differentiate \( -\frac{x^3}{3} \) to get \( -x^2 \).- Differentiate \( -\frac{1}{3}x \) to get \( -\frac{1}{3} \).Combining these gives \( x^{-2} - x^2 - \frac{1}{3} \), confirming our solution is correct.

Key Concepts

Antiderivative

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.

Integration Techniques

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.

Calculus Verification

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.