anonymous
  • anonymous
d/dt integral from 0,2x of (e^t+2t)dt is?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
the derivative of the integral? is that what it says?
amistre64
  • amistre64
\[\frac{d}{dt}(\int\limits_{0}^{2x} (e^2 +2t)dt)\]
anonymous
  • anonymous
yeah that's right.

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amistre64
  • amistre64
derivative and integrals are inverse functions; they cancel each other out...
amistre64
  • amistre64
your looking at a graph with a domain of [0,2x] with an equation of: y = e^t + 2t
amistre64
  • amistre64
does that make sense?
anonymous
  • anonymous
so far yes
amistre64
  • amistre64
is there a selection of answers to choose from?
anonymous
  • anonymous
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
  • amistre64
they appear to be using the (2x) as the variable; so my best guess is [A]. e^(2x) + 2(2x) = e^(2x) + 4x
anonymous
  • anonymous
In the answer sheet it says E, but I have no idea how they got it.
amistre64
  • amistre64
lol.....[E]? yeah, got no idea how theyd get that one....
anonymous
  • anonymous
lol.. i got the first part but idk how they got 8x
amistre64
  • amistre64
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
  • amistre64
....integrate the 2t up to t^2 :) typoed in my head lol
anonymous
  • anonymous
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
  • anonymous
does that sound about right?
amistre64
  • amistre64
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
  • amistre64
this appears to be the logic :)
anonymous
  • anonymous
yay :) Thank you!! lol

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