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Learn how the Fundamental Theorem of Calculus connects derivatives, integrals, and antiderivatives.

---

## The Bridge Between Derivatives and Integrals

The Fundamental Theorem of Calculus (FTC) is the central link between two ideas in calculus: **derivatives** and **integrals**. This theorem gives us a simpler way to calculate definite integrals without going through the long process of Riemann sum limits. The theorem is divided into two important parts.

These two parts complement each other: the first part shows that differentiation and integration are inverse operations, while the second part provides a practical way to calculate definite integrals using antiderivatives. This theorem is built upon the foundation of the **Mean Value Theorem for Integrals**, which guarantees the existence of a specific point within the integration interval.

## Differentiating an Integral Function

The first part of the FTC reveals that the processes of integration and differentiation are inverse operations. Formally, the theorem states:

```math
F(x) = \int_{a}^{x} f(t) \, dt \implies F'(x) = f(x)
```

This means that the derivative of a function defined as an integral is the function inside the integral itself.

> Imagine $$f(t)$$ is the rate at which water flows into a bucket at time $$t$$. Then, $$F(x)$$ is the total volume of water in the bucket at time $$x$$. This theorem says that the rate of change of the water's volume at that instant (<InlineMath math="F'(x)" />) is exactly equal to the flow rate at that instant ($$f(x)$$).

Visible text: > Imagine is the rate at which water flows into a bucket at time . Then, is the total volume of water in the bucket at time . This theorem says that the rate of change of the water's volume at that instant (<InlineMath math="F'(x)" />) is exactly equal to the flow rate at that instant ().

**Example:**

Find the derivative of $$F(x) = \int_{1}^{x} \sqrt{2+t} \, dt$$.

Visible text: Find the derivative of .

**Solution:**

Based on FTC Part One, we don't need to integrate the function. We simply replace the variable $$t$$ in the integrand with the upper limit $$x$$.

Visible text: Based on FTC Part One, we don't need to integrate the function. We simply replace the variable in the integrand with the upper limit .

```math
\frac{d}{dx} \left( \int_{1}^{x} \sqrt{2+t} \, dt \right) = \sqrt{2+x}
```

It's as simple as that. The derivative is the original function, evaluated at the upper limit.

## Evaluating Integrals with Antiderivatives

The second part, also known as the **Evaluation Theorem**, gives us a practical method for calculating the exact value of a definite integral. The theorem states:

```math
\int_{a}^{b} f(x) \, dx = F(b) - F(a)
```

Where $$F(x)$$ is the **antiderivative** (indefinite integral) of $$f(x)$$. This means that to find the area under the curve of $$f(x)$$ from $$a$$ to $$b$$, we just need to find its antiderivative and then calculate the difference in its value at the two limits.

Visible text: Where is the **antiderivative** (indefinite integral) of . This means that to find the area under the curve of from to , we just need to find its antiderivative and then calculate the difference in its value at the two limits.

To write this, we often use the bracket notation $$[F(x)]_{a}^{b}$$ which means $$F(b) - F(a)$$.

Visible text: To write this, we often use the bracket notation which means .

**Example** $$1$$:

Visible text: **Example** :

Find the value of $$\int_{1}^{3} x^2 \, dx$$.

Visible text: Find the value of .

**Solution:**

1.  **Find the antiderivative:** The antiderivative of $$f(x) = x^2$$ is $$F(x) = \frac{1}{3}x^3$$.
2.  **Evaluate at the limits:** Calculate $$F(3) - F(1)$$.

    <MathContainer>
    
    
    ```math
    \int_{1}^{3} x^2 \, dx = \left[ \frac{1}{3}x^3 \right]_{1}^{3}
    ```

    
    
    ```math
    = \left( \frac{1}{3}(3)^3 \right) - \left( \frac{1}{3}(1)^3 \right)
    ```

    
    
    ```math
    = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}
    ```

    </MathContainer>

Visible text: 1. **Find the antiderivative:** The antiderivative of is .
2. **Evaluate at the limits:** Calculate .

 <MathContainer>
 
 

 
 

 
 

 </MathContainer>

**Example** $$2$$:

Visible text: **Example** :

Find the value of $$\int_{1}^{5} (2x-1) \, dx$$.

Visible text: Find the value of .

**Solution:**

1.  **Find the antiderivative:** The antiderivative of $$f(x) = 2x - 1$$ is $$F(x) = x^2 - x$$.
2.  **Evaluate at the limits:** Calculate $$F(5) - F(1)$$.

    <MathContainer>
    
    
    ```math
    \int_{1}^{5} (2x-1) \, dx = \left[ x^2 - x \right]_{1}^{5}
    ```

    
    
    ```math
    = (5^2 - 5) - (1^2 - 1)
    ```

    
    
    ```math
    = (25 - 5) - (1 - 1)
    ```

    
    
    ```math
    = 20 - 0 = 20
    ```

    </MathContainer>

Visible text: 1. **Find the antiderivative:** The antiderivative of is .
2. **Evaluate at the limits:** Calculate .

 <MathContainer>
 
 

 
 

 
 

 
 

 </MathContainer>

The Fundamental Theorem of Calculus simplifies how we calculate areas and accumulations, turning limit problems into algebraic steps with antiderivatives.