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

- anonymous

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

- katieb

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- anonymous

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

- anonymous

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

- anonymous

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 \[\int^0_{-2} e^t,dt\]

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## More answers

- anonymous

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

- anonymous

Are you really being asked to use a Reimann sum?

- anonymous

yes

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

i have to go
good luck

- anonymous

thanks a lot!

- anonymous

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 \[e^3-e-6\]

- anonymous

could you please explain further?

- anonymous

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.

- anonymous

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.
\[\int^b_a c dx=(b-a)c\]

- anonymous

thanks a lot

- anonymous

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

- myininaya

see if this helps
its a geometric series

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- myininaya

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

- myininaya

almost there

##### 1 Attachment

- myininaya

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

- myininaya

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

- myininaya

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

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