The Fundamental Theorem, part one
Differentiation and integration undo one another, and the first part of the Fundamental Theorem of Calculus says exactly how. If \(f\) is continuous and
\[F(x) = \int_{a}^{x} f(t)\,dt\]
then \(F\) is differentiable and
\[\frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x)\]
The integral accumulates area; differentiating asks how fast the area is growing; and the answer is simply the height of the curve at the right-hand edge. Notice what happened to the constant lower limit \(a\) — nothing. It does not appear in the answer, because moving the starting point changes the total area by a constant, and constants differentiate to zero.
When the limit is a function of x
When the upper limit is a function of \(x\) rather than \(x\) itself, the chain rule joins in:
\[\frac{d}{dx}\int_{a}^{u(x)} f(t)\,dt = f(u(x))\,u'(x)\]
For example, \(\frac{d}{dx}\int_{0}^{x^{2}} \sin t\,dt = \sin(x^{2}) \cdot 2x\). Evaluate the integrand at the limit, then multiply by the derivative of that limit — the same "outside, then inside" pattern as any other composition.
Both limits varying
Split the integral at any convenient constant and apply the rule twice, with a sign for the lower limit:
\[\frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\,dt = f(u(x))\,u'(x) - f(v(x))\,v'(x)\]
The minus sign is not an extra rule — it comes from swapping the limits of the first piece, which flips the sign of an integral.
Four worked examples
- \(\frac{d}{dx}\int_{1}^{x} \frac{1}{t}\,dt = \frac{1}{x}\). The lower limit is a constant, so the answer is the integrand evaluated at \(x\). (This integral is \(\ln x\), whose derivative is indeed \(1/x\).)
- \(\frac{d}{dx}\int_{0}^{x^{3}} e^{t^{2}}\,dt = e^{x^{6}} \cdot 3x^{2}\). No antiderivative of \(e^{t^{2}}\) exists in elementary terms — and it is not needed.
- \(\frac{d}{dx}\int_{x}^{5} \cos t\,dt = -\cos x\). The variable is the lower limit, so the sign flips.
- \(\frac{d}{dx}\int_{x}^{x^{2}} \ln t\,dt = \ln(x^{2}) \cdot 2x - \ln(x) \cdot 1 = 4x\ln x - \ln x\), using \(\ln(x^{2}) = 2\ln x\).
Example 2 is the one worth remembering: the whole point of the theorem is that you never have to find the antiderivative.
When the integrand contains x too
When the integrand itself contains \(x\), differentiating under the integral sign needs the general Leibniz rule:
\[\frac{d}{dx}\int_{v(x)}^{u(x)} f(x, t)\,dt = f(x, u)\,u' - f(x, v)\,v' + \int_{v(x)}^{u(x)} \frac{\partial f}{\partial x}\,dt\]
The first two terms are the moving limits; the third accounts for the integrand changing shape as \(x\) moves. In a first-year course the third term is usually absent, because \(f\) depends only on \(t\).
Things to watch
- The calculator on this site differentiates expressions, not integrals. This page is the method; work the examples by hand.
- \(f\) must be continuous on the interval for the theorem to apply.
- A definite integral with two constant limits is just a number, so its derivative is zero.
- "Undo a derivative" is the opposite operation — antidifferentiation — and needs an integral calculator, not this one.
Questions
What is the derivative of an integral?
By the Fundamental Theorem of Calculus, \(\frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x)\) — the integrand evaluated at the upper limit.
What if the upper limit is a function of x?
Multiply by its derivative: \(\frac{d}{dx}\int_{a}^{u(x)} f(t)\,dt = f(u(x))\,u'(x)\). It is the chain rule.
What if x is the lower limit?
The sign flips: \(\frac{d}{dx}\int_{x}^{b} f(t)\,dt = -f(x)\), because swapping the limits of an integral negates it.
Do I need to find the antiderivative first?
No — that is the point of the theorem. \(\frac{d}{dx}\int_{0}^{x^{3}} e^{t^{2}}dt\) works out to \(3x^{2}e^{x^{6}}\) even though \(e^{t^{2}}\) has no elementary antiderivative.
What is the derivative of a definite integral with constant limits?
Zero. With both limits constant the integral is a fixed number, and constants differentiate to zero.