Question

I repost question :D (old question deleted, plz give a try to this question its hard :D )
USing the mean value theorem prove that: \[2e ^{-1/4}\le \int\limits_{0}^{2}e^{-r^2-r}dr \le 2e^2\]

Answer 1

mean value theorem of calculus :
\[\int\limits_{a}^{b}f(x)dx = f(x)(b-a)\]

Answer 2

doesn't look true to me

Answer 3

why?

Answer 4

(this question is taken from ODTU's (Thecnical University of Orta Dogu ) calculus final exam ^.^

Answer 5

\[2e^{-1/4}\approx1.5576\]
\[\int\limits_{0}^{2}e^{-r^2-r}dr\approx.545178\]

Answer 6

you see its true ^.^ but prove that by using mean value theorem :)

Answer 7

but i did a mistake at mean value theorem s0zi it should be .... =f(c)(b-a)

Answer 8

how is \[1.5576\le .545178\]?

Answer 9

is this 0.5....
or 5.4.....??

Answer 10

0.54517827464201

Answer 11

ahh k now i see -.-'

Answer 12

(lol @ me i red .54.... as 5.4 -.-')

Answer 13

ok i give up ...

Answer 14

I would give up to since\[2e ^{-1/4}\le \int\limits_{0}^{2}e^{-r^2-r}dr \le 2e^2\] is not true

Answer 15

*too