anonymous
  • anonymous
use the limit definition to evaluate the integral of (e^x-3)dx upper limit is 3, lower limit 1.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
let x-3 =t dx=dt hence our integral becomes \[\int\limits_{1}^{3}e ^{t}dt\]
anonymous
  • anonymous
the question is Evaluate using the limit definition, \[\int\limits_{1}^{3}(e^x-3)dx\]
anonymous
  • 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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anonymous
  • anonymous
oops Thought it was \[e^{x-3}\] not \[e^x-3\]
anonymous
  • anonymous
Are you really being asked to use a Reimann sum?
anonymous
  • anonymous
yes
anonymous
  • anonymous
is there a way to solve this without using reimann sum? do u have to take the natural log of e^x?
myininaya
  • myininaya
int(e^x-3) =int(e^x)-int(3) =e^x-3x+C C is a constant
anonymous
  • anonymous
hey could this be the final answer? u didnt use reimanns sum though
myininaya
  • 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
  • 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
  • myininaya
oops mistake =(e^3-3(3))-(e^1-3(1))
myininaya
  • myininaya
i have to go good luck
anonymous
  • anonymous
thanks a lot!
anonymous
  • 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
  • anonymous
could you please explain further?
anonymous
  • 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
  • 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
  • anonymous
thanks a lot
anonymous
  • anonymous
no problem but i still have no idea how to compute this as a reimann sum
myininaya
  • myininaya
see if this helps its a geometric series
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myininaya
  • myininaya
but maybe there is a mistake somewhere but i'm on the right track
myininaya
  • myininaya
almost there
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myininaya
  • 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
  • myininaya
\[=\lim_{u \rightarrow 0} \frac{1+u}{-1}=\lim_{u \rightarrow 0} (-1-u)=-1-0=-1\]
myininaya
  • myininaya
and (-1)(e-e^3)=e^3-e :) and we win!

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