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Second Fundamental Theorem of Calculus — Definition, Formula & Examples

The Second Fundamental Theorem of Calculus states that if you define a function as a definite integral with a variable upper bound, then the derivative of that function simply returns the original integrand evaluated at that upper bound.

If ff is continuous on an open interval containing aa, and F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt, then FF is differentiable and F′(x)=f(x)F'(x) = f(x) for every xx in that interval. In other words, differentiation undoes the accumulation process of integration.

Key Formula

ddx∫axf(t) dt=f(x)\frac{d}{dx}\int_a^x f(t)\,dt = f(x)
Where:
  • f(t)f(t) = A continuous function being integrated
  • aa = A fixed constant serving as the lower bound of integration
  • xx = The variable upper bound of integration

How It Works

Start with a continuous function f(t)f(t) and a fixed lower bound aa. Define a new function F(x)F(x) as the integral of ff from aa to xx. The theorem tells you that F′(x)=f(x)F'(x) = f(x) — you recover the original function by differentiating. When the upper bound is a function g(x)g(x) instead of just xx, apply the chain rule: ddx∫ag(x)f(t) dt=f(g(x))⋅g′(x)\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x).

Worked Example

Problem: Find the derivative of F(x)=∫1x(3t2+2) dtF(x) = \int_1^x (3t^2 + 2)\,dt.
Identify the integrand: The integrand is f(t)=3t2+2f(t) = 3t^2 + 2, and it is continuous everywhere. The lower bound is the constant a=1a = 1 and the upper bound is xx.
f(t)=3t2+2f(t) = 3t^2 + 2
Apply the theorem: By the Second Fundamental Theorem of Calculus, replace tt with xx in the integrand.
F′(x)=f(x)=3x2+2F'(x) = f(x) = 3x^2 + 2
Answer: F′(x)=3x2+2F'(x) = 3x^2 + 2

Why It Matters

This theorem is essential in AP Calculus AB/BC, where problems frequently ask you to differentiate accumulation functions. It also underpins physics concepts like finding velocity from a position integral, and it appears regularly on college calculus exams whenever the chain rule is combined with integral bounds.

Common Mistakes

Mistake: Forgetting the chain rule when the upper bound is a function like x2x^2 instead of plain xx.
Correction: If the upper limit is g(x)g(x), multiply by g′(x)g'(x): ddx∫ag(x)f(t) dt=f(g(x))⋅g′(x)\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x). For example, with an upper bound of x2x^2, you must multiply the result by 2x2x.