How do I simplify the expression Differentiating Integrals?

[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!

 

[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).

 

[AngelofMusic]

Thank you so much for your help!

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

And once that was fixed, I was able to follow your advice and differentiate this correctly.

(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

Thanks again for your help!

 

[HallsofIvy]

LalPlace's formula

d/dx (∫_g(x)^h(x)f(t,x)dt is

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