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Leibniz Integral Rule — Definition, Formula & Examples

The Leibniz Integral Rule tells you how to take the derivative of a definite integral when the integrand or the limits of integration depend on the variable you are differentiating with respect to.

If f(x,t)f(x, t) and ∂f∂t\frac{\partial f}{\partial t} are continuous on the relevant domain, and a(t)a(t) and b(t)b(t) are differentiable functions of tt, then ddt∫a(t)b(t)f(x,t) dx=f(b(t),t) b′(t)−f(a(t),t) a′(t)+∫a(t)b(t)∂f∂t(x,t) dx.\frac{d}{dt}\int_{a(t)}^{b(t)} f(x,t)\,dx = f\bigl(b(t),t\bigr)\,b'(t) - f\bigl(a(t),t\bigr)\,a'(t) + \int_{a(t)}^{b(t)} \frac{\partial f}{\partial t}(x,t)\,dx.

Key Formula

ddt∫a(t)b(t)f(x,t) dx=f(b(t),t) b′(t)−f(a(t),t) a′(t)+∫a(t)b(t)∂f∂t(x,t) dx\frac{d}{dt}\int_{a(t)}^{b(t)} f(x,t)\,dx = f\bigl(b(t),t\bigr)\,b'(t) - f\bigl(a(t),t\bigr)\,a'(t) + \int_{a(t)}^{b(t)} \frac{\partial f}{\partial t}(x,t)\,dx
Where:
  • tt = The parameter with respect to which you differentiate
  • xx = The variable of integration
  • f(x,t)f(x,t) = The integrand, which may depend on both x and t
  • a(t),b(t)a(t), b(t) = Lower and upper limits of integration, which may depend on t

How It Works

The rule has three pieces. The first term accounts for the upper limit changing: evaluate the integrand at the upper limit and multiply by its derivative. The second term does the same for the lower limit, but with a minus sign. The third term captures how the integrand itself changes with the parameter by integrating the partial derivative. When both limits are constants, the first two terms vanish and you simply move the derivative inside the integral.

Worked Example

Problem: Find ddt∫0t23xt dx\frac{d}{dt}\int_{0}^{t^2} 3xt\,dx.
Identify the pieces: Here f(x,t)=3xtf(x,t) = 3xt, a(t)=0a(t) = 0, b(t)=t2b(t) = t^2. So a′(t)=0a'(t) = 0, b′(t)=2tb'(t) = 2t, and ∂f∂t=3x\frac{\partial f}{\partial t} = 3x.
Apply the upper-limit term: Evaluate ff at x=t2x = t^2 and multiply by b′(t)b'(t).
f(t2,t)⋅b′(t)=3(t2)(t)⋅2t=6t4f(t^2, t)\cdot b'(t) = 3(t^2)(t)\cdot 2t = 6t^4
Apply the lower-limit term: Since a′(t)=0a'(t) = 0, this term is zero.
f(0,t)⋅a′(t)=0f(0, t)\cdot a'(t) = 0
Compute the integral term: Integrate ∂f∂t=3x\frac{\partial f}{\partial t} = 3x from 00 to t2t^2.
∫0t23x dx=3x22∣0t2=3t42\int_{0}^{t^2} 3x\,dx = \frac{3x^2}{2}\Big|_0^{t^2} = \frac{3t^4}{2}
Combine: Add all three pieces together.
6t4−0+3t42=15t426t^4 - 0 + \frac{3t^4}{2} = \frac{15t^4}{2}
Answer: ddt∫0t23xt dx=15t42\dfrac{d}{dt}\displaystyle\int_{0}^{t^2} 3xt\,dx = \dfrac{15t^4}{2}

Why It Matters

This rule appears frequently in physics and engineering when quantities like energy, probability, or flux are expressed as parameter-dependent integrals. It also underlies the technique known as "differentiation under the integral sign," a powerful method for evaluating difficult integrals that Richard Feynman famously championed.

Common Mistakes

Mistake: Forgetting the boundary terms when the limits depend on the parameter
Correction: If either limit is a function of tt, you must include the f(b(t),t) b′(t)f(b(t),t)\,b'(t) and f(a(t),t) a′(t)f(a(t),t)\,a'(t) terms. Only drop them when both limits are constants.

Related Terms