# I Derivative of A Def. Integral Equals Another Def. Integral?

Tags:
1. Nov 3, 2016

### terryphi

I'm going through the book "Elementry Differnetial Equations With Boundary Value Problems" 4th Eddition by William R. Derrick and Stanley I. Grossman.

On Page 138 (below) )

The authors take the derivative of a definite integral and end up with a definite integral plus another term. How did they do this?

2. Nov 4, 2016

### lurflurf

Use Leibniz integral rule
$$\frac{\mathrm{d}}{\mathrm{d}x}\int_{a(x)}^{b(x)} \! f(x,t) \, \mathrm{d}x=\frac{\mathrm{d}b(x)}{\mathrm{d}x}f(x,b(x))-\frac{\mathrm{d}a(x)}{\mathrm{d}x}f(x,a(x))+\int_{a(x)}^{b(x)} \! \frac{\mathrm{\partial}}{\mathrm{\partial}x} f(x,t) \, \mathrm{d}x$$
https://en.wikipedia.org/wiki/Leibniz_integral_rule

3. Nov 4, 2016

### andrewkirk

It is because they differentiate with respect to $x$ and there are at least three occurrences of $x$ in the formula that is being integrated: one as an upper integration limit, and two as function arguments in the integrand. There may be up to two more, being the $y_1,y_2$ in the denominator of the integrand. The excerpt does not make clear whether they also depend on $x$, so I will assume they do not, although they could, since there is a $y_1(x)$ and $y_2(x)$ in the numerator.

Let the function $A:\mathbb R^2\to\mathbb R$ be an antiderivative (indefinite integral) of the integrand with respect to $t$, ie $\frac{\partial }{\partial t}A(t,x)=a(t,x)$ where $a(t,x)$ is the integrand.

Then the definite integral is
$$A(x,x)-A(x_0,x)$$
Differentiating this wrt $x$ and using the total differential rule gives
\begin{align*}
\frac d{dx}\left(A(x,x)-A(x_0,x)\right)
&=D_1A(x,x)\frac{dx}{dx} + D_2A(x,x)\frac{dx}{dx}
-D_1A(x_0,x)\frac{dx_0}{dx}-D_2A(x_0,x)\frac{dx}{dx}\\
&=D_1A(x,x)\cdot 1 + D_2A(x,x)\cdot 1
-D_1A(x_0,x)\cdot 0-D_2A(x_0,x)\cdot 1\\
&=D_1A(x,x) + D_2A(x,x)-D_2A(x_0,x)
\end{align*}
where $D_kA$ indicates the partial derivative of function $A$ wrt its $k$th argument.
This is equal to
\begin{align*}
a(x,x) + D_u\bigg(A(x,u)-A(x_0,u)\bigg)\bigg|_{u=x}
&=
a(x,x) + D_u \int_{x_0}^x a(t,u)dt\bigg|_{u=x}
\end{align*}
where $D_u$ indicates partial differentiation wrt the variable $u$
In the second term we can, given certain mild assumptions, move the differentiation operator inside the integral sign to obtain:
\begin{align*}
a(x,x) + \int_{x_0}^x D_ua(t,u)dt\bigg|_{u=x}
=a(x,x) + \int_{x_0}^x D_xa(t,x)dt
\end{align*}

EDIT: Which is a derivation of the Leibniz integration rule referenced in the above post, which appeared on my screen when I posted this.

Last edited: Nov 4, 2016
4. Nov 4, 2016

### lurflurf

The y's in the denominator of the integral do not depend on x. After differentiating The t's in the denominator turn to x's.