# Deriving Generalized Leibniz’s Integral Rule

The general form of Leibniz’s Integral Rule with variable limits states:

Suppose $f(x, t)$ satisfies the condition stated previously for the basic form of Leibniz’s Rule (LR-1, see “A Semi-Rigorous Derivation of Leibniz’s Rule“) . In addition, $a(t), b(t)$ are defined and have continuous derivatives for $t_1\le t\le t_2.$ Then for $t_1\le t \le t_2,$

$\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx = f(b(t),t)\cdot b'(t) -f(a(t),t)\cdot a'(t)+ \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.\quad\quad\quad(1)$

(1) can be derived as a consequence of LR-1, the Multivariable Chain Rule, and the Fundamental Theorem of Calculus (FTC):

Clearly,

$\int\limits_{a(t)}^{b(t)}f(x, t)\;dx\quad\quad\quad(2)$

on the left side of (1) is a function of $t$.

Let

$u = a(t),\quad\quad\quad(3)$

$v = b(t),\quad\quad\quad(4)$

$w = t,\quad\quad\quad(5)$

(2) can be expressed as

$G(u,v,w) = \int\limits_{u}^{v}f(x,w)\;dx.$

Hence, by the chain rule,

$\frac{d}{d t}\int\limits_{u}^{v}f(x, w)\;dx = \frac{\partial}{\partial u}G(u,v,w)\cdot \frac{du}{dt} + \frac{\partial}{\partial v} G(u,v,w)\cdot \frac{dv}{dt} + \frac{\partial}{\partial w}G(u,v,w)\cdot \frac{dw}{dt}$

where

$\frac{\partial}{\partial u}G(u,v,w)=\frac{\partial}{\partial u}\int\limits_{u}^{v}f(x, w)\;dx$

$= \frac{\partial}{\partial u}\left(-\int\limits_{v}^{u}f(x, w)\;dx\right)$

$= -\frac{\partial}{\partial u}\int\limits_{v}^{u}f(x, w) \;dx$

$\overset{\textbf{FTC}}{=} -f(u, w)$

$\overset{(3), (5)}{=} -f(a(t), t),$

$\frac{\partial}{\partial v}G(u,v,w)=\frac{\partial}{\partial v}\int\limits_{u}^{v}f(x, w)\;dx \overset{\textbf{FTC}}{=}f(v, w) \overset{(4), (5)}{=} f(b(t), t)$

and,

$\frac{\partial}{\partial w}G(u,v,w)=\frac{\partial}{\partial w}\int\limits_{u}^{v}f(x, w)\;dx \overset{\textbf{LR-1}}{=} \int\limits_{u}^{v}\frac{\partial}{\partial w}f(x, w)\;dx\overset{(3), (4), (5)}{=} \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.$

It follows that

$\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx =-f(a(t),t)\cdot a'(t) + f(b(t),t)\cdot b'(t) + \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx,$

i.e.,

$\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx = f(b(t),t)\cdot b'(t) -f(a(t),t)\cdot a'(t)+ \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.$