Leibniz Integral Rule
Separate moving-endpoint and integrand contributions to the derivative of a signed polynomial integral.
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Differentiating an integral with two sources of change
Let F(x) be the integral from a to kx of t2 + xt. The lower bound a and multiplier k stay constant while x varies. The upper endpoint moves, and the value of the integrand at each fixed t changes too. Leibniz's rule accounts for both effects.
The endpoint contributes k[(kx)2 + x(kx)]. The partial derivative of the integrand is t, whose integral contributes [(kx)2 − a2]/2. Add these terms to obtain dF/dx. Direct integration gives F(x) = (k3/3 + k2/2)x3 − a2x/2 − a3/3, and differentiating this cubic independently gives the same answer.
Worked example: a moving endpoint and a changing curve
With a = 0, k = 1 and x = 1, the integral is 1/3 + 1/2 = 5/6. The endpoint has speed 1 and integrand value 2, so its contribution is 2. The interior contribution is the integral of t from 0 to 1, which is 1/2. The total derivative is therefore 5/2, even though F itself is only 5/6.
| a, k, x | F | Endpoint | Interior | dF/dx |
|---|---|---|---|---|
| 0, 1, 1 | 5/6 | 2 | 1/2 | 5/2 |
| 1, 1, 1 | 0 | 2 | 0 | 2 |
| 0, 2, 1 | 14/3 | 12 | 2 | 14 |
| 0, 1, −1 | −5/6 | 2 | 1/2 | 5/2 |
Experiment with the two contributions
Open the interactive experiment and enter these ordinary decimal values. Read the four numerical cards after each edit; the curve and endpoint markers use the same current inputs. The graph can be panned independently of the calculation.
- Set Lower bound a to 0, Upper-bound multiplier k to 1 and Parameter x to 1. Check Integral F(x) = 0.8333333333 and Total derivative dF/dx = 2.5.
- Change Lower bound a to 1. The interval collapses and F becomes 0, but the moving-boundary contribution and total derivative remain 2.
- Set Lower bound a to 0 and Upper-bound multiplier k to 2. The endpoint moves twice as fast; check the contributions 12 and 2, totaling 14.
- Set Lower bound a to 1 and Upper-bound multiplier k to 0. Check F = −0.8333333333 and derivative −0.5: only the integrand changes.
- Restore Lower bound a to 0 and Upper-bound multiplier k to 1, then set Parameter x to 0. All four numbers become 0.
- Set Parameter x to −1. The interval is Reversed, F = −0.8333333333 and the derivative is 2.5. The calculation keeps the bound orientation.
- Drag the graph, then select Reset view. Parameter x stays −1 and the numerical results remain unchanged.
- Clear Parameter x to see the error and hidden dependent results. Enter 1 to restore the calculation, or select Reset inputs to restore all defaults.
Scope and limitations
This is a finite, smooth polynomial example. The lower endpoint and k are fixed parameters of the derivative, not functions of x. Negative x and reversed bounds are allowed: swapping bounds changes an integral's sign, so the model never replaces it with an absolute area. General parameter-dependent improper integrals require additional hypotheses before exchanging a derivative and an integral. Decimal readouts are rounded, while the displayed formulas specify the exact mathematics.
Reference
The general moving-boundary differentiation formula is presented in the Princeton PlasmaWiki explanation of the Leibniz Integral Rule. The polynomial, calculated examples and input bounds here specialize that formula to an experiment whose two derivative contributions can be checked directly.
Experience it interactively
Adjust parameters, observe in real time, and build deep intuition with Riano’s interactive Leibniz Integral Rule module.
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