Question

d/dt integral from 0,2x of (e^t+2t)dt is?

Answer 1

the derivative of the integral? is that what it says?

Answer 2

\[\frac{d}{dt}(\int\limits_{0}^{2x} (e^2 +2t)dt)\]

Answer 3

yeah that's right.

Answer 4

derivative and integrals are inverse functions; they cancel each other out...

Answer 5

your looking at a graph with a domain of [0,2x] with an equation of: y = e^t + 2t

Answer 6

does that make sense?

Answer 7

so far yes

Answer 8

is there a selection of answers to choose from?

Answer 9

yeah they are:
A) e^2x+4x
B)e^2x+4x-1
C) e^2x+4x^2-1
D)2e^2x+4x
E)2e^2x+8x

Answer 10

they appear to be using the (2x) as the variable; so my best guess is [A].

e^(2x) + 2(2x) = e^(2x) + 4x

Answer 11

In the answer sheet it says E, but I have no idea how they got it.

Answer 12

lol.....[E]? yeah, got no idea how theyd get that one....

Answer 13

lol.. i got the first part but idk how they got 8x

Answer 14

we can try this; go ahead and integrate first to:

e^(t) + t ; now fill in the interval [0,2x]

e^(2x) + 2x -e^(0)

D{t} [e(2x) +2x - 1]

2e^(2x) + 2 ??

Answer 15

....integrate the 2t up to t^2 :) typoed in my head lol

Answer 16

nvm i think i got it...
you know how ur supposed to find the derivative and since there is no 2 you add it. Then if you distribute it you're gonna get 2e^t+4t. and then when i plug in the 2x i get E

Answer 17

does that sound about right?

Answer 18

e^(t) + t^2 ; now fill in the interval [0,2x]

e^(2x) + (2x)^2 -e^(0)

D{t} [e(2x) +4x^2 - 1]

2e^(2x) + 8x

Answer 19

this appears to be the logic :)

Answer 20

yay :)
Thank you!! lol