Introduction
When you learn calculus, two big topics come up.
- Differentiation: finding the slope of a graph (the instantaneous rate of change).
- Integration: finding the area under a curve.
At first these seem completely separate. What could slope and area possibly have to do with each other?
Here is the one-sentence answer up front — "how fast the area is growing" is exactly the value of the function at that instant. In other words, if you write the accumulated area as a function, its derivative is the original function. This is the heart of the Fundamental Theorem of Calculus.
In this post, we will derive why, step by step, from intuition all the way to the formula.
Let's build an "area function"
Given a curve \(y = f(t)\), let us bundle up the area under the curve from the starting point \(t = 0\) to \(t = x\) and call it the area function \(A(x)\).
$$A(x) = \int_0^x f(t)\,dt$$As \(x\) moves a little to the right, the shaded region grows wider. \(A(x)\) is the function that tracks that area.
Now, what happens if we differentiate this \(A(x)\) with respect to \(x\)?
Intuition — the speed at which the area grows
Suppose we move just a tiny bit, from \(x\) to \(x + h\). How much did the area increase?
$$A(x+h) - A(x) = \int_x^{x+h} f(t)\,dt$$This value is almost equal to the area of a small rectangle whose width is \(h\) and whose height is roughly \(f(x)\).
$$\int_x^{x+h} f(t)\,dt \approx f(x) \cdot h \quad (h \text{ is very small})$$Dividing both sides by \(h\):
$$\frac{A(x+h) - A(x)}{h} \approx f(x)$$As we let \(h \to 0\), this approximation becomes exact, and the left-hand side is precisely the definition of the derivative.
$$A'(x) = \lim_{h \to 0} \frac{A(x+h)-A(x)}{h} = f(x)$$The derivative of the area function equals the original function \(f(x)\). This is the Fundamental Theorem of Calculus, Part 1 (FTC Part 1).
To put it intuitively again — when you slip in a thin slice of width \(h\) on the right, the area of that slice is "height \(f(x)\) × width \(h\)." Divide the speed at which the area grows by the width, and only the height \(f(x)\) remains. A perfectly natural result.
Checking it with the formula
As an example, let us take \(f(t) = (t-1)^2 + 1\). This function has its minimum value of 1 at \(t = 1\), and it is always positive.
$$A(x) = \int_0^x \bigl[(t-1)^2 + 1\bigr]\,dt$$Computing it directly:
$$A(x) = \left[\frac{(t-1)^3}{3} + t\right]_0^x = \frac{(x-1)^3}{3} + x - \left(\frac{(-1)^3}{3} + 0\right) = \frac{(x-1)^3}{3} + x + \frac{1}{3}$$Simplifying:
$$A(x) = \frac{x^3}{3} - x^2 + 2x$$Now let us differentiate \(A(x)\).
$$A'(x) = x^2 - 2x + 2 = (x-1)^2 + 1 = f(x)$$We got exactly \(f(x)\). The Fundamental Theorem of Calculus holds.
Let us check at a few points.
| \(x\) | \(A(x)\) | \(A'(x)\) | \(f(x)\) | Match? |
|---|---|---|---|---|
| 0 | 0 | 2 | 2 | ✓ |
| 1 | \(4/3\) | 1 | 1 | ✓ |
| 2 | \(8/3\) | 2 | 2 | ✓ |
| 3 | 6 | 5 | 5 | ✓ |
Try it yourself
In the interactive below, drag the x slider. The blue area in the left graph is \(A(x)\), and the right graph shows that same \(A(x)\) growing. As you push the slider to the right, you can confirm that the speed at which the area grows (the orange figure) matches \(f(x)\) at that instant (the blue figure).
Try moving x slowly from 1 to 3. The higher the function, the more steeply the area piles up; the lower the function (near x=1), the more gently the area accumulates.
Fundamental Theorem of Calculus, Part 2 — computing area with an antiderivative
If FTC Part 1 says "the derivative of the area function = the original function," then Part 2 flips that around for practical use.
If you know a function \(F(x)\) satisfying \(F'(x) = f(x)\) (an antiderivative, or indefinite integral), then:
$$\int_a^b f(x)\,dx = F(b) - F(a)$$Checking with the example above, \(F(x) = x^3/3 - x^2 + 2x\), and:
$$\int_0^3 f(x)\,dx = F(3) - F(0) = \left(9 - 9 + 6\right) - 0 = 6$$This figure matches exactly the \(A(3) = 6\) displayed in the interactive when you set x to 3.
Thanks to this, instead of "chopping the area into tiny pieces and adding them up" with a Riemann sum, you only need to find an antiderivative and compute the difference of its values at the two endpoints. It shows how the fact that differentiation and integration are operations in opposite directions actually works in practical computation.
Why "opposite" — an addition-and-subtraction analogy
Just as addition and subtraction are inverse operations of each other, integration and differentiation are inverse operations too.
- Add 3 to a number and then subtract 3, and you return to the original number.
- Integrate a function to build an area function, then differentiate it again, and you return to the original function.
And yet the fact that these two operations are "opposites" is not obvious from the start. Integration is defined by a Riemann sum (chopping into infinitely tiny pieces and adding them up), while differentiation is defined by a limit. That two operations starting from such different points turn out to be inverses of each other is precisely the heart of the Fundamental Theorem of Calculus — and historically the great insight that Newton and Leibniz discovered independently.
Key takeaways
| Question | Answer |
|---|---|
| What is the area function A(x)? | The area under the curve from the starting point to x |
| What do you get differentiating A(x)? | f(x) — you return to the original function |
| Why? | Slice area of width h ≈ f(x)·h → speed = f(x) |
| What does FTC Part 2 mean? | Compute a definite integral with an antiderivative F: F(b) − F(a) |
| What is the relationship between differentiation and integration? | Inverse operations of each other, like addition and subtraction |
Studying this with AI
AI can explain the Fundamental Theorem of Calculus from several angles. Ask specifically about wherever you get stuck.
Examples of useful questions:
- "Explain why differentiating \(\int_0^x t^2\,dt\) gives \(x^2\)."
- "Explain intuitively why a Riemann sum and an antiderivative give the same value."
- "Compute \(\int_1^3 (2x+1)\,dx\) using the antiderivative method."
Wrapping up
The reason integration and differentiation are opposite operations is not because of formula rules. It starts from the intuition that "how fast the area is growing" equals the value of the function at that instant, and once you take the limit, it follows inevitably.
The Fundamental Theorem of Calculus ties together two ideas that look utterly different — area and slope — into one, letting you use the differentiation tools you have learned so far directly for computing integrals.
Related reading — Read first: Why Is the Derivative the "Instantaneous Slope"? · Why Is the Chain Rule a "Multiplication"? → Up next: What Are the du and dv in Integration by Parts, Really? · Draw the graphs yourself: Desmos · Wolfram Alpha