## anonymous 5 years ago Integral from 1 to 2 of ( v (e-e inverse) dv

1. anonymous

$\int\limits_{1}^{2}v (e-e^-1)$

2. anonymous

easy

3. amistre64

e is a constant; and so is 1/e

4. anonymous

2 ( e-e^-1 )

5. anonymous

wait thats wrong lol

6. anonymous

take the constant out

7. amistre64

might as well ask what is the integral of 4v dv :)

8. anonymous

integral v from 2 to 1

9. anonymous

(1/2) [ 4-1] = 3/2 so $\frac{3}{2} (e - e ^{-1})$

10. anonymous

11. amistre64

$F(v)=\frac{v^2(e-e^{-1})}{2} |_{1}^{2}$

12. amistre64

e - (1/e) = (e^2 - 1)/e right?

13. anonymous

So don't treat v as a constant

14. anonymous

its dv on the end

15. amistre64

v is the variable here

16. anonymous

Ok

17. anonymous

18. amistre64

4.7008047745751485263396882938069 -1.1752011936437871315849220734517 -------------------------------------- 3.525603580931361394754766220355 is what i get :)

19. anonymous

:|

20. amistre64

F(2) - F(1)

21. anonymous

stuff like that is what puts people off maths ^ :|

22. anonymous

no one cares about the numbers, especially to 20 decimal places , the final number doesnt matter , its how you go the number that matters Whenever you have e's or $\pi$ in the answer then the answer should be given as exact value

23. anonymous

are you in kindergarten amistre lol

24. amistre64

i gave the exact and the approximate; or so i thought....

25. amistre64

i even gave a reason of "how" it is obtained.... right?

26. anonymous

your like the guys that memorise pi to 1000's of demical places

27. amistre64

lol.... 7 digits on pi :) e = 2.71828182845905... is that best i got with that one :)

28. anonymous

never give an answer to 20decimal places, makes you lokok like an idiot, 2 ( maybe 3 ) max , no one cares bout the rest. When a person that is new to maths sees something really long and unneccessary like that it is sure to put them off

29. anonymous

*look like .....

30. amistre64

i dont have to include decimals to look like an idiot; im a pro at that regardless :) and i figured that if they are asking calculus questions then they arent all that new to math...