Fundamental Theorem of Calculus

The Bridge Between Derivatives and Integrals

The Fundamental Theorem of Calculus (FTC) is the central pillar of calculus, magnificently connecting two seemingly different concepts: derivatives and integrals. This theorem provides us with a much simpler and more powerful method for calculating definite integrals, without needing to go through the lengthy 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:

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

As a simple analogy, 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 ($F'(x)$) is exactly equal to the flow rate at that instant ($f(x)$).

Example:

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

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 $x$.

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:

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.

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

Example 1:

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

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 $F(3) - F(1)$.

   $$\int_{1}^{3} x^2 \, dx = \left[ \frac{1}{3}x^3 \right]_{1}^{3}$$

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

   $$= \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$$

Example 2:

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

Solution:

1. Find the antiderivative: The antiderivative of $f(x) = 2x - 1$ is $F(x) = x^2 - x$.

2. Evaluate at the limits: Calculate $F(5) - F(1)$.

   $$\int_{1}^{5} (2x-1) \, dx = \left[ x^2 - x \right]_{1}^{5}$$

   $$= (5^2 - 5) - (1^2 - 1)$$

   $$= (25 - 5) - (1 - 1)$$

   $$= 20 - 0 = 20$$

The Fundamental Theorem of Calculus fundamentally simplifies how we calculate areas and accumulations, turning them from complex limit problems into straightforward algebraic processes using antiderivatives.