Question

If g(x) = the integral from 0 to x^2 of (t-4)dt, what does g'(x) equal? Using the fundamental theorem of calculus, I think the answer would be (2x-4), but I'm not sure. Also, what is the largest value of g(x) in this interval? I set 2x=4 and solve x=2, and said this is the maximum, but once again not sure.

Answer 1

wrong

Answer 2

if g(x)=integral from 0 to x2 f(t)dt
then g'(x)=f(x2)*2x

Answer 3

you mean g'(x) = f(x^2)*2x?

Answer 4

How can one prove this?

Answer 5

it is newton leibniz formula

Answer 6

google it for the proper form ...i mean generalised one

Answer 7

Alright, thank you!

Answer 8

no problem

Answer 9

In general:
\[g(x) = \int_a^{h(x)} f(t) dt = F(h(x))-F(a) \]
Then:
\[\frac{dg}{dt} = \frac{d}{dx}[F(h(x))] - \frac{d}{dx}[F(a)] = f(h(x))\frac{dh}{dx}-0 \]