Question

Need to find G ' (2)

Answer 1

I don't know what to do with the integral part

Answer 2

\[G(x)=x^2+\int\limits_{-2}^{2x}f(t)dt\]

Answer 3

I am thinking that I have to find the derivatives from -2 to 4, in that case it turns into 0
but I am not sure

Answer 4

G'(x) = 2x + f(2x)
G'(2) = 2(2) + f(4) = 4 + 0 = 4

Answer 5

the derivative of the integral is zero, but the details are hazy.

Answer 6

so what I doing it correct by looking at the derivatives of the integral? of was I suppose to do f(2x)

Answer 7

it might be f(2x) - f(-2) which gives f(4) - f(-2) = 0 - 0 = 0
but like I said, it's a bit hazy, and I will have to google it.

Answer 8

@aum care to elaborate for us?

Answer 9

Fundamental Theorem of Calculus (part 1)

\(\Large F(x) = \int_a^xf(t)dt \\ \Large
F'(x) = f(x)
\)

Answer 10

it's just f(2x)

Answer 11

http://mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html

Answer 12

Thanks, I guess I get confused because the drawing is involved

Answer 13

You are welcome.

Answer 14

the drawing gives you the value of f(4)

Answer 15

thanks, I will bookmark that page for future reference @phi

Answer 16

You may find this video useful too: https://www.youtube.com/watch?v=PGmVvIglZx8

Answer 17

thanks aum

Answer 18

Oh, I forgot to include the 2 above.

G'(x) = 2x + 2 * f(2x)
G'(2) = 2(2) + 2 * f(4) = 4 + 2* 0 = 4