Question

PLEASE HELP! URGENT! MEDAL! @ganeshie8

Answer 1

@ganeshie8 @hartnn

Answer 2

\(\Large \dfrac{d}{dx} \int \limits_{f(x)}^{g(x)} h(t)dt = [h(t)]_{f(x)}^{g(x)} = h[g(x)] -h[f(x)]\)

Answer 3

so just plug in 7x^2 first in your function and then 2

Answer 4

huh?

Answer 5

I think we don't to worry about the bottom 2

Answer 6

g(x) = 7x^2?

Answer 7

Yes

Answer 8

so thats because plugging in 2 will result in constant and derivative of constant is 0 ?

Answer 9

yeah, we may use this :
\[\dfrac{d}{dx}\int\limits_{a}^{g(x)}h(t)\, dt = h(g(x))*g'(x)\]

Answer 10

oh yes! we do need g'(x) too

Answer 11

So do \(int_{0}^{7x^2}\sqrt{2+\cos^{3}(7x^2)}\)?

Answer 12

times the derivative of 7x^2, which is 14x?

Answer 13

PS. that int is supposed to be an integral. don't know why it didn't come out right

Answer 14

just
\(\sqrt{2+\cos^{3}(7x^2)} \times (14x)\)

Answer 15

okay.

Answer 16

So is the answer just \(\Large14x\sqrt{2+\cos^3(7x^2)}\)? Or do I need to do u-substitution or something like that?

Answer 17

yes, thats it.
no substitution.

note this down:
\(\dfrac{d}{dx}\int\limits_{a}^{g(x)}h(t)\, dt = h(g(x))\times g'(x)\)

Answer 18

Okay, thank you both for your help. :)