 Proof that
 ∫𝑓'(𝑥) 𝑑𝑥 = 𝑓(𝑥)

To prove that:

\[
\int f'(x) \, dx = f(x)
\]

we will use **the Fundamental Theorem of Calculus**, which is the cornerstone of the relationship between differentiation and integration.

### Step-by-step proof:

1. **Fundamental Theorem of Calculus (Part 1)**:
   The first part of the Fundamental Theorem of Calculus states:
   
   \[
   \frac{d}{dx} \left( \int_{a}^{x} f'(t) \, dt \right) = f'(x)
   \]
   This means that the derivative of an integral of a function's derivative returns the original function's derivative, \(f'(x)\).

2. **Antiderivative and indefinite integral**:
   The second part of the theorem tells us that if \(F(x)\) is an antiderivative of \(f'(x)\), then:
   
   \[
   \int f'(x) \, dx = f(x) + C
   \]
   where \(C\) is the constant of integration. In this case, \(F(x) = f(x)\), because taking the derivative of \(f(x)\) gives \(f'(x)\).

### Conclusion:
Thus, integrating \(f'(x)\) with respect to \(x\) yields:

\[
\int f'(x) \, dx = f(x) + C
\]

This proves that the indefinite integral of the derivative of a function is the function itself, up to an arbitrary constant of integration.

Would you like more details or have any questions?

Here are 5 related questions:
1. Why do we add the constant \(C\) when integrating?
2. What is the Fundamental Theorem of Calculus?
3. How do definite and indefinite integrals differ?
4. What happens if you integrate a constant?
5. Can every function \(f(x)\) be integrated?

**Tip**: When performing indefinite integrals, always remember to include the constant of integration \(C\).

Proof that ∫𝑓'(𝑥) 𝑑𝑥 = 𝑓(𝑥)

Understand the proof of the integral of f'(x) with respect to x equaling f(x), explained using the Fundamental Theorem of Calculus. This step-by-step explanation covers key concepts like differentiation, integration, and the role of constants. Ideal for students in Grades 11-12 or undergraduate calculus courses.

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