## anonymous 5 years ago d/dt integral from 0,2x of (e^t+2t)dt is?

1. amistre64

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

2. amistre64

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

3. anonymous

yeah that's right.

4. amistre64

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

5. amistre64

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

6. amistre64

does that make sense?

7. anonymous

so far yes

8. amistre64

is there a selection of answers to choose from?

9. 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

10. amistre64

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

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

12. amistre64

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

13. anonymous

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

14. 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 ??

15. amistre64

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

16. 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

17. anonymous

18. 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

19. amistre64

this appears to be the logic :)

20. anonymous

yay :) Thank you!! lol