Fundamental Theorem of Calculus Part 1: Derivative of an Integral

The fundamental theorem of calculus, part 1 (FTC 1) says that if you build a function by accumulating area under a continuous curve, the derivative of that function is the curve itself. It is the first half of the link between derivatives and integrals.

Statement of FTC part 1

Let f be continuous on [a, b], and define the accumulation function

Then F is continuous on [a, b], differentiable on (a, b), and

In Leibniz notation, \frac{d}{dx}\int_a^x f(t)\,dt = f(x). The variable t is a dummy variable of integration, and x is the upper limit.

What FTC 1 means

F(x) is the signed area under y = f(t) from a to x. Nudge x a little to the right by h, and the new area is a thin strip of width h and height about f(x). So

and the approximation becomes exact as h \to 0. The rate at which area accumulates equals the height of the curve.

Why continuity matters

Let f(t) = 0 for t < 0 and f(t) = 1 for t \ge 0, and let F(x) = \int_{-1}^x f(t)\,dt. The accumulated area is

which has a corner at x = 0. So F'(0) does not exist, and certainly does not equal f(0) = 1. The integral still makes sense, but at the jump the conclusion F' = f fails. Away from the jump, where f is continuous, F' = f still holds.

Worked examples

Example 1: the basic form.

No integration needed, and none possible in elementary terms. Replace t with x and you are done.

Example 2: upper limit is a function of x (chain rule). Find \frac{d}{dx}\int_0^{x^2}\cos t\,dt.

Step 1. Let u = x^2. Then the integral is G(u) = \int_0^u \cos t\,dt, and by FTC 1, G'(u) = \cos u.

Step 2. By the chain rule, \frac{d}{dx}G(x^2) = \cos(x^2)\cdot 2x.

Check: the integral equals \sin(x^2), whose derivative is 2x\cos(x^2).

Example 3: both limits depend on x. Find \frac{d}{dx}\int_x^{2x} e^{t^2}\,dt.

Step 1. Split at a constant: \int_x^{2x} = \int_0^{2x} - \int_0^{x}.

Step 2. Differentiate each piece with FTC 1 and the chain rule:

Example 4: a lower limit with x. Flip the limits first: \frac{d}{dx}\int_x^5 \sin(t^2)\,dt = \frac{d}{dx}\left(-\int_5^x \sin(t^2)\,dt\right) = -\sin(x^2).

Common mistakes

1. Forgetting the chain rule factor. \frac{d}{dx}\int_0^{x^2}\cos t\,dt is 2x\cos(x^2), not \cos(x^2).
2. Plugging in the lower limit too. For \int_a^x f(t)\,dt with constant a, the answer is just f(x). The constant lower limit only shifts F by a constant.
3. Forgetting the sign when x is in the lower limit. Swapping the limits introduces a minus sign.
4. Integrating first when it is impossible. Integrands like e^{t^2} have no elementary antiderivative, but FTC 1 still differentiates the integral instantly.

Why FTC 1 matters

FTC 1 proves that every continuous function has an antiderivative, namely \int_a^x f(t)\,dt. That fact is what makes FTC part 2 work, turning definite integrals into simple subtraction. It also defines important functions such as \ln x = \int_1^x \frac{1}{t}\,dt. For the full story of both parts, read the fundamental theorem of calculus guide, and see Riemann sums for the definition of the integral it relies on.