Fundamental Theorem of Calculus: Integration

Hello! Welcome to your next lesson. In the previous module, we finished our deep dive into optimization, learning how to find the maximum and minimum values of functions, even when faced with constraints. That whole area of calculus revolved around differentiation—the study of rates of change.

Today, we pivot to the other major pillar of calculus: integration. If differentiation is about finding a function's slope at a point, integration is about accumulating its values over an interval. Think of it this way: if you know the velocity of a car at every instant (rate of change), integration allows you to find the total distance it has traveled (accumulated quantity).

Our goal for this lesson is to evaluate basic definite and indefinite integrals using the Fundamental Theorem of Calculus. This theorem is the centerpiece of the topic because it provides a profound and powerful link between differentiation and integration, essentially showing they are inverse processes.

1. What is an Integral? The Idea of Accumulation

At its heart, integration is a process of summing up infinitely many, infinitesimally small things. The most common way to visualize this is by calculating the area under the graph of a function.

Imagine you have a function, say, f(x), and you want to find the area between its curve and the x-axis, from a starting point x=a to an ending point x=b. How would you do it? A clever way is to slice that area into a huge number of very thin rectangles, calculate the area of each one, and add them all up. As you make the rectangles infinitely thin, the sum of their areas approaches the true area under the curve.

This idea is brilliantly explained in the following video.

Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus

To build a strong intuition for what an integral represents, please watch the first part of this video from the 'Essence of calculus' series by 3Blue1Brown. It connects the physical problem of finding distance from velocity to the geometric problem of finding the area under a curve.

Watch from the beginning until 09:23. Pay close attention to: How the problem of finding total distance from a changing velocity is set up. The strategy of approximating the area under the velocity curve using thin rectangles. The introduction of the integral symbol \int_a^b f(t) dt as the precise value that this sum of rectangular areas approaches.

As the video explains, this process of finding the "area under the curve" gives us two main concepts:

Definite Integral: This is written as $\int_{a}^{b} f (x) d x$ . It represents a specific numerical value—the signed area under the curve of $f (x)$ from the lower limit $a$ to the upper limit $b$ . "Signed" means that area below the x-axis counts as negative.
Indefinite Integral: This is written as $\int f (x) d x$ (without limits). It doesn't ask for a number, but for a function. Specifically, it asks: "What function, when differentiated, gives us $f (x)$ ?" This function is called the antiderivative.

2. The Fundamental Theorem of Calculus: The Grand Connection

You might wonder, "Calculating the areas of infinite rectangles sounds complicated. Is there an easier way?" The answer is a resounding yes, and it is one of the most beautiful results in all of mathematics.

The Fundamental Theorem of Calculus (FTC) provides the crucial link between differentiation and integration.

The video you just watched continues to explain this very idea. It shows that if you define a function that gives the area under a curve up to a variable endpoint, the derivative of that "area function" is simply the original function itself.

Fundamental Theorem of Calculus (Part 1) Explained — This infographic illustrates Part 1 of the FTC, which states that differentiation and integration are inverse processes. If g(x) is the integral (area function) of f(x), then the derivative of g(x) is f(x).

This leads directly to the practical part of the theorem, often called FTC Part 2, which gives us a stunningly simple method to calculate definite integrals.

Fundamental Theorem of Calculus (FTC) - Part 2 — This slide illustrates Part 2 of the FTC. To compute the definite integral of f(x) from a to b, you just need to find its antiderivative F(x), and then compute F(b) - F(a).

In simple terms, the theorem states:

To evaluate the definite integral $\int_{a}^{b} f (x) d x$ :

First, find an antiderivative of $f (x)$ . Let's call it $F (x)$ . This means finding a function $F (x)$ such that $F^{'} (x) = f (x)$ .

Then, simply calculate $F (b) - F (a)$ .

This completely bypasses the need to sum up infinite rectangles!

Calculus I - Computing Definite Integrals

This article from Paul's Online Math Notes provides a concise and clear explanation of this computational process. Please read the introduction to see the formal statement of the theorem.

Read the first part of the section, down to just before 'Example 1'. Focus on the formal statement of the 'Fundamental Theorem of Calculus, Part II' and the discussion about why the constant of integration (+C) is not needed for definite integrals.

3. Finding Antiderivatives: The Power Rule in Reverse

The FTC tells us that the key to integration is finding the antiderivative. Let's start with the most common functions you'll encounter: polynomials.

You know the power rule for differentiation: the derivative of $x^{n}$ is $n x^{n - 1}$ . To find the antiderivative, we just reverse this process.

Instead of subtracting 1 from the exponent, we add 1.
Instead of multiplying by the original exponent, we divide by the new exponent.

So, the indefinite integral of $x^{n}$ is:

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C

The + C is called the constant of integration. It's there because the derivative of any constant is zero. So, if $F (x)$ is an antiderivative of $f (x)$ , then so is $F (x) + 5$ , $F (x) - 100$ , or $F (x) + C$ for any constant $C$ . For indefinite integrals, this +C is essential.

This next video provides many clear, step-by-step examples of how to find antiderivatives and use them to evaluate definite integrals.

Calculus 1 - Integration & Antiderivatives

This video by The Organic Chemistry Tutor is excellent for learning the mechanics of integration. Watch the following segments to master the basic rules.

Please watch these three segments: 00:00 - 07:52 (Sections: Power Rule & Polynomials): This covers the power rule for integration and how to apply it term-by-term to polynomials. 17:04 - 23:39 (Section: Definite vs. Indefinite Integrals): This part is crucial. It shows exactly how to use an antiderivative to evaluate a definite integral using the formula F(b) - F(a). It also reinforces why the +C is not needed for definite integrals.

4. Putting It All Together: A Worked Example

Let's walk through an example from start to finish.
Problem: Evaluate $\int_{1}^{2} (y^{2} + y^{- 2}) d y$ .

This is Example 1b from the Paul's Online Math Notes resource (LINK).

Find the indefinite integral (antiderivative):
We apply the power rule to each term separately.
- For $y^{2}$ , the antiderivative is $\frac{y ^{2 + 1}}{2 + 1} = \frac{y ^{3}}{3}$ .
- For $y^{- 2}$ , the antiderivative is $\frac{y ^{- 2 + 1}}{- 2 + 1} = \frac{y ^{- 1}}{- 1} = - \frac{1}{y}$ .
  So, our antiderivative function is $F (y) = \frac{y ^{3}}{3} - \frac{1}{y}$ .
Apply the Fundamental Theorem of Calculus:
We need to compute $F (b) - F (a)$ , which is $F (2) - F (1)$ .
- $F (2) = \frac{2 ^{3}}{3} - \frac{1}{2} = \frac{8}{3} - \frac{1}{2}$
- $F (1) = \frac{1 ^{3}}{3} - \frac{1}{1} = \frac{1}{3} - 1$
Simplify:
$(\frac{8}{3} - \frac{1}{2}) - (\frac{1}{3} - 1) = \frac{8}{3} - \frac{1}{2} - \frac{1}{3} + 1 = \frac{7}{3} + \frac{1}{2} = \frac{14 + 3}{6} = \frac{17}{6}$

Test your understanding!

Evaluate the definite integral $\int_{0}^{3} (2 x^{2} - 4 x) d x$ .

Show answer

Find the antiderivative of $f (x) = 2 x^{2} - 4 x$ .
Using the power rule in reverse:
$F (x) = 2 (\frac{x ^{3}}{3}) - 4 (\frac{x ^{2}}{2}) = \frac{2}{3} x^{3} - 2 x^{2}$ .
Evaluate at the limits using the FTC: $F (3) - F (0)$ .
- $F (3) = \frac{2}{3} (3)^{3} - 2 (3)^{2} = \frac{2}{3} (27) - 2 (9) = 18 - 18 = 0$ .
- $F (0) = \frac{2}{3} (0)^{3} - 2 (0)^{2} = 0$ .
Calculate the result:
$F (3) - F (0) = 0 - 0 = 0$ .

The value of the definite integral is 0. This means the positive area and negative area under the curve from x=0 to x=3 cancel each other out perfectly!

An Important Caveat: Continuity

The Fundamental Theorem of Calculus has one important requirement: the function $f (x)$ must be continuous on the interval $[a, b]$ . If there is a discontinuity (like a division by zero) within the interval, you cannot apply the theorem directly.

For example, consider the integral $\int_{- 1}^{2} \frac{1}{y ^{2}} d y$ . The function $f (y) = \frac{1}{y ^{2}}$ has a division-by-zero error at $y = 0$ , which is inside the interval $[- 1, 2]$ . Therefore, we cannot compute this integral using the methods we've learned today. Always check for continuity before integrating!

Conclusion

Great work today! You've taken your first and most important step into the world of integral calculus.

Key Takeaways:

An indefinite integral $\int f (x) d x$ asks for the antiderivative of a function, which is a family of functions $F (x) + C$ .
A definite integral $\int_{a}^{b} f (x) d x$ asks for a numerical value representing the signed area under a curve.
The Fundamental Theorem of Calculus provides the bridge: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , where $F$ is an antiderivative of $f$ .
We can find antiderivatives of basic functions by reversing the rules of differentiation, such as the power rule.

Preview of the Next Lesson:

Now that we've seen how integration is the reverse of differentiation, we are going to loop back to strengthen our differentiation skills. In the next lesson, we will cover the essential techniques for differentiating more complex functions: the chain rule, product rule, and quotient rule. Mastering these will not only complete your toolkit for differentiation but will also be crucial for tackling more advanced integration problems in the future.