Fundamental theorem of calculus

The fundamental theorem of calculus connects the two big ideas of calculus: derivatives and integrals are inverse operations. It comes in two parts, and you use both all the time.

By the Calcist team · Updated October 29, 2026 · About a 7-minute read

The two parts

Part 1: if F(x) = ∫ₐˣ f(t) dt, then F′(x) = f(x)
Part 2: ∫ₐᵇ f(x) dx = F(b) − F(a), where F is any antiderivative of f

Both parts assume f is continuous on the interval. Part 1 says that accumulating area and then differentiating gets you back where you started. Part 2 is the practical one: to find the exact area under a curve, you don't need to add up rectangles; you need an antiderivative.

Part 2: evaluating definite integrals

Example 1. ∫₁³ x² dx. An antiderivative of x² is x³/3. So the integral is 3³/3 − 1³/3 = 9 − 1/3 = 26/3 ≈ 8.667.

Example 2. ∫₀^π sin x dx. An antiderivative is −cos x. So the integral is −cos π − (−cos 0) = 1 + 1 = 2. The area under one arch of the sine curve is exactly 2.

Example 3. ∫₁^e (1/x) dx = ln e − ln 1 = 1 − 0 = 1.

Any antiderivative works, because the + C cancels: (F(b) + C) − (F(a) + C) = F(b) − F(a). That's why you can leave it off in definite integrals.

Part 1: the derivative of an integral

Part 1 lets you differentiate a function defined as an integral without ever finding the integral.

Example 4. If F(x) = ∫₀ˣ √(1 + t³) dt, then F′(x) = √(1 + x³). There's no elementary antiderivative here, and you don't need one.

Example 5 (with the chain rule). If the upper limit is a function g(x), multiply by g′(x): d/dx ∫₀^(x²) cos t dt = cos(x²) · 2x = 2x cos(x²).

Example 6 (variable lower limit). Flip the limits, which changes the sign: d/dx ∫ₓ⁵ t³ dt = d/dx (−∫₅ˣ t³ dt) = −x³.

Expression	Derivative
∫ₐˣ f(t) dt	f(x)
∫ₐ^g(x) f(t) dt	f(g(x)) · g′(x)
∫_h(x)^b f(t) dt	−f(h(x)) · h′(x)
∫_h(x)^g(x) f(t) dt	f(g(x)) · g′(x) − f(h(x)) · h′(x)

The net change theorem

Part 2 has a useful reading: integrating a rate of change gives the total change. If v(t) is velocity, ∫ₐᵇ v(t) dt is the change in position from time a to time b. For example, with v(t) = 3t² + 2 meters per second, the displacement over the first 4 seconds is [t³ + 2t]₀⁴ = 64 + 8 = 72 m. Divide a net change by the length of the interval and you get an average rate of change.

Why it's true, briefly

Let A(x) be the area under f from a to x. Increase x by a small h: the extra area is a thin strip about f(x) wide and h long, so A(x + h) − A(x) ≈ f(x)·h. Divide by h and let h → 0: A′(x) = f(x). That's Part 1. Then A is an antiderivative of f, and any other antiderivative F differs by a constant, so ∫ₐᵇ f = A(b) − A(a) = F(b) − F(a). That's Part 2. The mean value theorem makes this argument precise.

Common mistakes

Subtracting in the wrong order. It's F(top) − F(bottom).
Forgetting the chain rule factor g′(x) when the upper limit isn't just x.
Using Part 2 across a discontinuity. ∫₋₁¹ (1/x²) dx is not [−1/x]₋₁¹ = −2: the function blows up at 0, so the integral is improper and actually diverges.

Evaluate definite integrals with steps in the integral calculator, see how rectangle sums approach the exact value in the Riemann sum calculator, and differentiate with the derivative calculator. For products, see integration by parts.

Sources

Every worked example on this page was checked independently with SymPy.

Questions people ask

What does the fundamental theorem of calculus say?

Part 1: the derivative of ∫ₐˣ f(t) dt is f(x). Part 2: ∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F of f. Together they say differentiation and integration undo each other.

How do you use the fundamental theorem of calculus to evaluate an integral?

Find an antiderivative F, then subtract its value at the lower limit from its value at the upper limit. For ∫₁³ x² dx, that's 27/3 − 1/3 = 26/3.

What is the derivative of an integral with x² as the upper limit?

Plug x² into the integrand and multiply by 2x. For example, d/dx ∫₀^(x²) cos t dt = 2x cos(x²).

Why don't you add + C in definite integrals?

Because it cancels: (F(b) + C) − (F(a) + C) = F(b) − F(a).

Does the theorem need f to be continuous?

Yes, as usually stated. At a point where f blows up, the integral is improper and has to be handled with limits.