What is this question asking?

Question

Integration:

(The interval for both is 0 to x.)
Let H(x) = *interval* (g(x)-g(t))f(t)dt.

Show that H'(x) = g'(x) *interval* f(t)dt.

(Sorry, it's hard to type this sort of problem out.)

Am I just trying to prove the two are equal?

Thanks!

Answer 1

I'm assuming that "interval" means the interGRal. Using "S" to mean the intergral sign, here's the proof. You have

H(x) = S (g(x)-g(t))f(t)dt
H(x) = Sg(x)f(t)dt - Sg(t)f(t)dt

Since g(x) isn't a function of t, it factors out as a constant:

H(x) = g(x)* Sf(t)dt - Sg(t)f(t)dt

Take the derivative of both sides with respect to t. This means the integrals become what's inside, and since the intervals go from 0 to t you just replace t with x. Notice you need to use the chain rule on the first term:

H'(x) = (g'(x)* Sf(t)dt) + g(x)f(x)) - g(x)f(x)
H'(x) = g'(x)* Sf(t)dt)

And that's what we were trying to prove. I'm also assuming that you can just let the deriviatives cancel out the integrals like this, but I might be missing a technicality.

Answer 2

I'm going to assume that you're using *interval* for the integral sign.

Then yes, the question is to show that H'(x) is equal to g'(x) *interval* f(t) dt.

Answer 3

they are not equal. you are going to have to differentiate H(x) to get to the exact form of H'(x).