In short

The Fundamental Theorem of Calculus has two parts. Part 1 says that if you define a function by integrating f from a fixed point up to x, the derivative of that function is f(x) — differentiation undoes integration. Part 2 says that you can evaluate any definite integral \int_a^b f(x)\,dx by finding an antiderivative F and computing F(b) - F(a). Together, they reveal that the area problem and the tangent-slope problem are inverse operations.

There are two great problems in calculus. One is ancient: given a curve, find the area underneath it. The other is medieval: given a curve, find the slope of the tangent at a point.

These look like completely different questions. One is about summing (adding up thin strips of area). The other is about shrinking (taking a ratio of tiny changes and sending them to zero). One is integration. The other is differentiation.

The Fundamental Theorem of Calculus says: these two operations are inverses of each other. Differentiating undoes integrating, and integrating undoes differentiating. The theorem that proves this is not just a technical result — it is the single idea that unifies the entire subject. Without it, integration and differentiation would be two separate theories. With it, they are two faces of the same coin.

Part 1: Differentiation undoes integration

Define a new function G as follows. Pick a starting point a. For any x \geq a, let

G(x) = \int_a^x f(t)\,dt

This function records the total signed area under f from a up to x. As x moves to the right, G(x) accumulates more area.

The question is: how fast is G growing? What is G'(x)?

Think about what happens when you nudge x to x + h, for a small positive h:

G(x + h) - G(x) = \int_a^{x+h} f(t)\,dt - \int_a^x f(t)\,dt = \int_x^{x+h} f(t)\,dt

The difference in accumulated area is the area of the thin strip from x to x + h. That strip has width h and height approximately f(x) (since h is small and f is continuous, the function does not change much over this tiny interval). So:

G(x+h) - G(x) \approx f(x) \cdot h

Divide by h:

\frac{G(x+h) - G(x)}{h} \approx f(x)

Take the limit as h \to 0:

G'(x) = \lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = f(x)

The rate at which area accumulates equals the height of the curve. That is Part 1.

Fundamental Theorem of Calculus — Part 1

Let f be continuous on [a, b]. Define G(x) = \int_a^x f(t)\,dt for x \in [a, b]. Then G is differentiable on (a, b), and

G'(x) = f(x)

In Leibniz notation: \dfrac{d}{dx}\!\int_a^x f(t)\,dt = f(x).

What Part 1 is really saying. Integration creates a new function — the area accumulator G. Part 1 says the derivative of this accumulator is the original function f. In other words, the antiderivative of f is constructed by integration. Differentiation undoes the integration.

The proof, made rigorous

The intuitive argument above has one gap: the claim that \int_x^{x+h} f(t)\,dt \approx f(x) \cdot h. Here is the precise version.

Since f is continuous on the closed interval [x, x+h] (assuming h > 0), f attains a minimum value m and a maximum value M on this interval. So:

m \cdot h \leq \int_x^{x+h} f(t)\,dt \leq M \cdot h

Divide by h (which is positive):

m \leq \frac{G(x+h) - G(x)}{h} \leq M

As h \to 0^+, both m and M approach f(x) (by continuity of f). By the squeeze theorem:

\lim_{h \to 0^+} \frac{G(x+h) - G(x)}{h} = f(x)

The same argument works for h \to 0^- (with the inequality directions adjusted). So G'(x) = f(x).

The proof uses three facts: f is continuous, continuous functions on closed intervals have max and min (the Extreme Value Theorem), and the squeeze theorem for limits. Each of these has its own rigorous proof, but together they give a clean, airtight argument.

Part 2: The evaluation theorem

Part 1 tells you that G(x) = \int_a^x f(t)\,dt is an antiderivative of f. But what if you already know a different antiderivative — call it F, so F'(x) = f(x)?

Since G and F are both antiderivatives of f, they differ by a constant: G(x) = F(x) + C for some constant C.

Now evaluate at x = a: G(a) = \int_a^a f(t)\,dt = 0 (the area from a to a is zero). So 0 = F(a) + C, which means C = -F(a).

Therefore: G(x) = F(x) - F(a).

Evaluate at x = b:

\int_a^b f(t)\,dt = G(b) = F(b) - F(a)

That is Part 2.

Fundamental Theorem of Calculus — Part 2 (Evaluation Theorem)

Let f be continuous on [a, b], and let F be any antiderivative of f (i.e., F'(x) = f(x)). Then

\int_a^b f(x)\,dx = F(b) - F(a)

What Part 2 is really saying. To compute the area under a curve from a to b, you don't need to do any Riemann sums. Find any function whose derivative is f, plug in b and a, and subtract. The connection between integration (summing areas) and differentiation (computing slopes) is so tight that knowing one immediately gives you the other.

This is why all the integration techniques you have been learning — substitution, parts, partial fractions — matter. They are tools for finding antiderivatives. Once you have the antiderivative, Part 2 converts it into the definite integral with a single subtraction.

Part 1 in action: differentiating an integral

Part 1 is not just a theoretical result — it appears directly in problems.

Example 1: Differentiating a variable-limit integral

Find \dfrac{d}{dx}\displaystyle\int_1^x (3t^2 - 2t + 5)\,dt.

Step 1. Identify the structure.

The integral has a variable upper limit x and a fixed lower limit 1. The integrand is f(t) = 3t^2 - 2t + 5, which is continuous everywhere.

Why: this is exactly the setup of Part 1. G(x) = \int_1^x f(t)\,dt, and Part 1 says G'(x) = f(x).

Step 2. Apply Part 1 directly.

\frac{d}{dx}\int_1^x (3t^2 - 2t + 5)\,dt = 3x^2 - 2x + 5

Why: replace t with x in the integrand. The derivative of the area accumulator is the integrand evaluated at the upper limit.

Step 3. Verify by computing the integral explicitly.

\int_1^x (3t^2 - 2t + 5)\,dt = \bigl[t^3 - t^2 + 5t\bigr]_1^x = (x^3 - x^2 + 5x) - (1 - 1 + 5) = x^3 - x^2 + 5x - 5

Differentiate: \frac{d}{dx}(x^3 - x^2 + 5x - 5) = 3x^2 - 2x + 5. Same answer.

Why: the explicit computation confirms Part 1. But Part 1 is faster — it skips the antiderivative step entirely.

Step 4. What if the upper limit is not x but x^2?

\frac{d}{dx}\int_1^{x^2} (3t^2 - 2t + 5)\,dt

By the chain rule: \frac{d}{dx}\int_1^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x).

Here g(x) = x^2, so g'(x) = 2x:

= \bigl(3(x^2)^2 - 2(x^2) + 5\bigr) \cdot 2x = (3x^4 - 2x^2 + 5) \cdot 2x

Why: the chain rule accounts for the fact that the upper limit is not simply x but a function of x. The derivative "brings down" the extra factor g'(x).

Result: \dfrac{d}{dx}\displaystyle\int_1^x (3t^2-2t+5)\,dt = 3x^2-2x+5. With upper limit x^2: (3x^4-2x^2+5) \cdot 2x.

The red curve is $G(x) = \int_1^x (3t^2-2t+5)\,dt = x^3-x^2+5x-5$. The black curve is $G'(x) = 3x^2-2x+5$. At every point, the slope of the red curve equals the height of the black curve — Part 1 of the Fundamental Theorem, made visible.

Part 2 in action: computing a definite integral

Example 2: Using Part 2 to evaluate a definite integral

Compute \displaystyle\int_0^{\pi/2} \cos x\,dx.

Step 1. Find an antiderivative of \cos x.

F(x) = \sin x

Why: \frac{d}{dx}[\sin x] = \cos x, so \sin x is an antiderivative of \cos x.

Step 2. Evaluate F at the limits.

F(\pi/2) = \sin(\pi/2) = 1, \qquad F(0) = \sin(0) = 0

Why: substitute the upper and lower limits into the antiderivative.

Step 3. Subtract.

\int_0^{\pi/2} \cos x\,dx = 1 - 0 = 1

Why: Part 2 says \int_a^b f(x)\,dx = F(b) - F(a). The area under one arch of the cosine curve is exactly 1 — a clean, exact answer.

Step 4. Interpret geometrically. The curve y = \cos x starts at height 1 (when x = 0), decreases smoothly, and reaches 0 (when x = \pi/2). The area trapped between the curve and the x-axis over this quarter-period is exactly 1 square unit.

Why: this is a reassuring sanity check. The region is roughly a quarter-ellipse, fitting inside a rectangle of width \pi/2 \approx 1.57 and height 1, so the area should be less than 1.57. An area of 1 fits.

Result: \displaystyle\int_0^{\pi/2} \cos x\,dx = 1.

The curve $y = \cos x$ from $x = 0$ to $x = \pi/2$. The region between the curve and the $x$-axis has area exactly $1$. The Fundamental Theorem converts this geometric question into arithmetic: $\sin(\pi/2) - \sin(0) = 1 - 0 = 1$.

Common confusions

Going deeper

The Fundamental Theorem is the central result of calculus, and the material above covers the statement, the proof, and the standard applications. If you are satisfied, you can stop here. Below, we explore what happens when the hypotheses fail, and the theorem's role in a broader context.

When both limits are variable

A common exam question: find \frac{d}{dx}\int_{x^2}^{x^3} \sin(t^2)\,dt.

Split using a constant (say 0):

\int_{x^2}^{x^3} \sin(t^2)\,dt = \int_0^{x^3} \sin(t^2)\,dt - \int_0^{x^2} \sin(t^2)\,dt

Now differentiate each piece using Part 1 with the chain rule:

\frac{d}{dx}\int_0^{x^3} \sin(t^2)\,dt = \sin((x^3)^2) \cdot 3x^2 = 3x^2\sin(x^6)
\frac{d}{dx}\int_0^{x^2} \sin(t^2)\,dt = \sin((x^2)^2) \cdot 2x = 2x\sin(x^4)

Subtract: \frac{d}{dx}\int_{x^2}^{x^3} \sin(t^2)\,dt = 3x^2\sin(x^6) - 2x\sin(x^4).

The general rule: for \int_{g(x)}^{h(x)} f(t)\,dt:

\frac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)

This is Leibniz's integral rule — a direct consequence of Part 1 and the chain rule applied to both limits.

Functions without elementary antiderivatives

Part 2 says: find an antiderivative and subtract. But what if no elementary antiderivative exists?

Consider \int_0^1 e^{-x^2}\,dx. The function e^{-x^2} is continuous, well-behaved, and the integral certainly exists (it represents a finite area under a bell curve). But there is no antiderivative expressible in terms of elementary functions — no combination of polynomials, exponentials, logarithms, trigonometric functions, and their inverses.

This does not mean the integral is impossible to compute. It means Part 2 cannot be applied in its simple form. The integral is evaluated using:

Part 1 still applies: G(x) = \int_0^x e^{-t^2}\,dt is a perfectly good function, and G'(x) = e^{-x^2}. The theorem is not limited by elementary functions — only the evaluation shortcut of Part 2 is.

The deeper meaning

The Fundamental Theorem says that integration and differentiation are inverse operations. But they are inverse in a nuanced way.

\frac{d}{dx}\int_a^x f(t)\,dt = f(x) — differentiating an integral recovers f. This is exact.

\int_a^x F'(t)\,dt = F(x) - F(a) — integrating a derivative recovers F, up to a constant. The constant F(a) is the "initial condition."

This is analogous to multiplication and division being inverse operations, but a/b \cdot b = a while a \cdot b / b = a only if you don't lose information in the intermediate step. Integration "remembers" the accumulated change but "forgets" the starting value. The constant of integration C in indefinite integrals is exactly this forgotten information.

Where this leads next