Question

Integral from 1 to 2 of ( v (e-e inverse) dv

Answer 1

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

Answer 2

easy

Answer 3

e is a constant; and so is 1/e

Answer 4

2 ( e-e^-1 )

Answer 5

wait thats wrong lol

Answer 6

take the constant out

Answer 7

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

Answer 8

integral v from 2 to 1

Answer 9

(1/2) [ 4-1] = 3/2

so \[\frac{3}{2} (e - e ^{-1})\]

Answer 10

^ answer

Answer 11

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

Answer 12

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

Answer 13

So don't treat v as a constant

Answer 14

its dv on the end

Answer 15

v is the variable here

Answer 16

Ok

Answer 17

So which answer is correct

Answer 18

4.7008047745751485263396882938069
-1.1752011936437871315849220734517
--------------------------------------

3.525603580931361394754766220355 is what i get :)

Answer 19

:|

Answer 20

F(2) - F(1)

Answer 21

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

Answer 22

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

Answer 23

are you in kindergarten amistre lol

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

*look like .....

Answer 30

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...