Question

use the limit definition to evaluate the integral of (e^x-3)dx upper limit is 3, lower limit 1.

Answer 1

let x-3 =t
dx=dt
hence our integral becomes
\[\int\limits_{1}^{3}e ^{t}dt\]

Answer 2

the question is Evaluate using the limit definition, \[\int\limits_{1}^{3}(e^x-3)dx\]

Answer 3

maybe a mistake here. If you change from x - 3 to t you have to change the limits of integration.
if x = 1 then t = 1-3 = -2
and
if x = 3 then t = 3-3 = 0
so integral should be \[\int^0_{-2} e^t,dt\]

Answer 4

oops
Thought it was \[e^{x-3}\] not \[e^x-3\]

Answer 5

Are you really being asked to use a Reimann sum?

Answer 6

yes

Answer 7

is there a way to solve this without using reimann sum? do u have to take the natural log of e^x?

Answer 8

int(e^x-3)
=int(e^x)-int(3)
=e^x-3x+C
C is a constant

Answer 9

hey could this be the final answer? u didnt use reimanns sum though

Answer 10

well you also have the limits too
this is alot easier than reimann sums
it would be easy to do the reimann sums on int(3)
but int(e^x) not sure never even tried to use riemann sum for that one

Answer 11

int(e^x-3,x=3..1)
=(e^x-3x,x=3..1)
=(e^3-3(3))-(e^1-1(1))
thats the answer
you can simplify of course

Answer 12

oops mistake
=(e^3-3(3))-(e^1-3(1))

Answer 13

i have to go
good luck

Answer 14

thanks a lot!

Answer 15

holy moly. i see no way to compute this reimann sum because we do not have a summation formula for this. the integral is easy enough:
\[\int^3_1e^x,dx=e^3-3\]
\[\int^3_1 3,dx=6\]
since it is a constant and the length of the path is 2.
so answer is \[e^3-e-6\]

Answer 16

could you please explain further?

Answer 17

sure. We want a definite integral. Easy way is to find the anti-derivative, plug in the upper limit, the lower limit, and subtract. It is easy to find the anti-derivative of \[e^x\] since it is its own derivative. so
\[\int e^x,dx = e^x\]
evaluate at 3 get \[e^3\]
evaluate at 1 get \[e\]
subtract and get \[e^3-e\]

now 3 is a constant. a horizontal line. So the integral is just base times hight. the hight is 3, the base is the length from 1 to 3 which is 2, and 3*2=6. I am ignoring the "-" because i am just going to subtract.

Answer 18

i guess hight is spelled with an e as height. not that literate today.
you could also find this integral by taking the anti-derivative:
\[\int 3 dx = 3x\]
plug in 3 get 9. plug in 1 get 3. subtract get 6. but this is a waste of time, because the integral of a constant is always constant times length of path.
\[\int^b_a c dx=(b-a)c\]

Answer 19

thanks a lot

Answer 20

no problem but i still have no idea how to compute this as a reimann sum

Answer 21

see if this helps
its a geometric series

Answer 22

but maybe there is a mistake somewhere but i'm on the right track

Answer 23

almost there

Answer 24

use l'hospital's rule and you have \[\lim_{u \rightarrow 0} \frac{ue^u}{1-e^{u}}=\lim_{u \rightarrow 0} \frac{e^u+ue^{u}}{-e^{u}}\]

Answer 25

\[=\lim_{u \rightarrow 0} \frac{1+u}{-1}=\lim_{u \rightarrow 0} (-1-u)=-1-0=-1\]

Answer 26

and (-1)(e-e^3)=e^3-e
:)
and we win!