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anonymous

  • 5 years ago

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

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  1. amistre64
    • 5 years ago
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    the derivative of the integral? is that what it says?

  2. amistre64
    • 5 years ago
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    \[\frac{d}{dt}(\int\limits_{0}^{2x} (e^2 +2t)dt)\]

  3. anonymous
    • 5 years ago
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    yeah that's right.

  4. amistre64
    • 5 years ago
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    derivative and integrals are inverse functions; they cancel each other out...

  5. amistre64
    • 5 years ago
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    your looking at a graph with a domain of [0,2x] with an equation of: y = e^t + 2t

  6. amistre64
    • 5 years ago
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    does that make sense?

  7. anonymous
    • 5 years ago
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    so far yes

  8. amistre64
    • 5 years ago
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    is there a selection of answers to choose from?

  9. anonymous
    • 5 years ago
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    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

  10. amistre64
    • 5 years ago
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    they appear to be using the (2x) as the variable; so my best guess is [A]. e^(2x) + 2(2x) = e^(2x) + 4x

  11. anonymous
    • 5 years ago
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    In the answer sheet it says E, but I have no idea how they got it.

  12. amistre64
    • 5 years ago
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    lol.....[E]? yeah, got no idea how theyd get that one....

  13. anonymous
    • 5 years ago
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    lol.. i got the first part but idk how they got 8x

  14. amistre64
    • 5 years ago
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    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 ??

  15. amistre64
    • 5 years ago
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    ....integrate the 2t up to t^2 :) typoed in my head lol

  16. anonymous
    • 5 years ago
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    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

  17. anonymous
    • 5 years ago
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    does that sound about right?

  18. amistre64
    • 5 years ago
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    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

  19. amistre64
    • 5 years ago
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    this appears to be the logic :)

  20. anonymous
    • 5 years ago
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    yay :) Thank you!! lol

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