# Differentiating Integrals

1. Dec 3, 2003

### AngelofMusic

Hello,

My question has to do with differentiating an integral. We are given:

$$f(x)=1/2\int_{0}^{x} (x-t)^2 g(t)dt$$

And we are asked to prove that:

$$f'(x)=x\int_{0}^{x}g(t)dt - \int_{0}^{x}tg(t)dt$$

My Solution:

I expanded $$(x-t)^2$$ into $$x^2-2xt+t$$ and then expanded

$$f(x)=x^2/2\int_{0}^{x}g(t)dt - x\int_{0}^{x}tg(t)dt + \int_{0}^{x}t/2g(t)dt$$

Then I differentiated using the product rule:

$$f'(x)=x\int_{0}^{x}g(t)dt + g(t)x^2/2 - \int_{0}^{x}tg(t)dt - xtg(t) + t/2g(t)$$

This is close to what they want me to prove, but I have these extra terms:

$$g(t)x^2/2-xtg(t)+t/2g(t)$$

Which is basically $$(t-x)^2/2g(t)$$.

How do I eliminate that term? Does it evaluate to 0 somehow?

Thanks for the help!

Last edited: Dec 3, 2003
2. Dec 3, 2003

### mathman

You can work out the details yourself, but what you overlooked, when you took the derivative, is that the integrals themselves are functions of x, since the upper limit is x. The derivatives are the integrands, g(t) or tg(t).

3. Dec 3, 2003

### AngelofMusic

Thank you so much for your help!

I noticed a mistake in my expansion of $$(x-t)^2$$

$$(x-t)^2=x^2-2xt+t^2$$

So,

$$f(x)=x^2/2\int_{0}^{x}g(t)dt - x\int_{0}^{x}tg(t)dt + \int_{0}^{x}t^2/2g(t)dt$$

$$f'(x)=x\int_{0}^{x}g(t)dt + g(x)x^2/2 - \int_{0}^{x}tg(t)dt - x^2g(x)+x^2/2g(x)$$

And then the $$x^2/2g(x)-x^2g(x)+x^2/2g(x)$$ cancel each other out, and

$$f'(x)=x\int_{0}^{x}g(t)dt - \int_{0}^{x}tg(t)dt$$

4. Dec 3, 2003

### HallsofIvy

Staff Emeritus
LalPlace's formula

d/dx (&int;g(x)h(x)f(t,x)dt is

&int;g(x)h(x)&partial;f(t,x)/&partial;xdt + dg/dx f(x,g(x))- df/dx f(x,h(x))