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

Question

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

yeah that's right.

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

does that make sense?

anonymous · Answer

so far yes

amistre64 · Answer

is there a selection of answers to choose from?

anonymous · Answer

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

amistre64 · Answer

they appear to be using the (2x) as the variable; so my best guess is [A].

e^(2x) + 2(2x) = e^(2x) + 4x

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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  ??

amistre64 · Answer

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

anonymous · Answer

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

anonymous · Answer

does that sound about right?

amistre64 · Answer

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

amistre64 · Answer

this appears to be the logic :)

anonymous · Answer

yay :)
Thank you!! lol