# Homework Help: ∂∫f(x,t)dt =? ∫∂f(x,t)dt

1. Dec 14, 2011

Just wondering what I've done wrong, assumed, messed up etc... with the following derivation of Leibniz's differentiating under the integral rule,

$$\int_{a(x)}^{b(x)}f(x,t)dt \ = \ F(x,b(x)) \ - \ F(x,a(x))$$

$$\frac{d}{dx} \ \int_{a(x)}^{b(x)}f(x,t)dt \ = \ \frac{d}{dx} \ [ \ F(x,b(x)) \ - \ F(x,a(x)) \ ]$$

$$\frac{d}{dx} \ \int_{a(x)}^{b(x)}f(x,t)dt \ = \ \frac{ \partial}{ \partial x}F(x,b(x)) \ + \frac{ \partial F}{ \partial b} \frac{db}{dx} \ - \ \frac{ \partial}{ \partial x}F(x,a(x)) \ - \ \frac{ \partial F}{ \partial a}\frac{da}{dx}$$

$$\frac{d}{dx} \ \int_{a(x)}^{b(x)}f(x,t)dt \ = \ \frac{ \partial F}{ \partial b} b'(x) \ - \frac{ \partial F}{ \partial a}a'(x) + \frac{ \partial}{ \partial x}[F(x,b(x)) \ - F(x,a(x))]$$

$$\frac{d}{dx} \ \int_{a(x)}^{b(x)}f(x,t)dt \ = \ f(x,b(x))b'(x) \ - f(x,a(x))a'(x) + \frac{ \partial}{ \partial x}[F(x,b(x)) \ - F(x,a(x))]$$

$$\frac{d}{dx} \ \int_{a(x)}^{b(x)}f(x,t)dt \ = \ f(x,b(x))b'(x) \ - f(x,a(x))a'(x) + \frac{ \partial}{ \partial x}[\int_{a(x)}^{b(x)}f(x,t)dt]$$

Now, assuming every step holds up to here the question is whether:

$$\frac{ \partial}{ \partial x} \ \int_{a(x)}^{b(x)}f(x,t)dt \ = \ \int_{a(x)}^{b(x)}\frac{ \partial}{ \partial x}f(x,t)dt$$

I've been told it doesn't, & I mean I could plug in values to check this but my concern is
what's wrong with everything I've done up to this and why this wrong step led to this.

Assuming it's not wrong I don't see why you can't justify putting the partial operator inside the integrand by using:

$$\frac{ \partial}{ \partial x} \ \sum_{i=1}^{n}f(x,t_i)\Delta t_i \ = \ \sum_{i=1}^{n}\frac{ \partial}{ \partial x}f(x,t_i)\Delta t_i$$

& taking limits so that we're integrating?
I can derive this another way I'm just curious about the apparent flaws with this idea because the last line is so close to the actual statement of the theorem, thanks.

Last edited: Dec 14, 2011
2. Dec 14, 2011