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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    let x-3 =t dx=dt hence our integral becomes \[\int\limits_{1}^{3}e ^{t}dt\]

  2. anonymous
    • 5 years ago
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    the question is Evaluate using the limit definition, \[\int\limits_{1}^{3}(e^x-3)dx\]

  3. anonymous
    • 5 years ago
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    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 \[\int^0_{-2} e^t,dt\]

  4. anonymous
    • 5 years ago
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    oops Thought it was \[e^{x-3}\] not \[e^x-3\]

  5. anonymous
    • 5 years ago
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    Are you really being asked to use a Reimann sum?

  6. anonymous
    • 5 years ago
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    yes

  7. anonymous
    • 5 years ago
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    is there a way to solve this without using reimann sum? do u have to take the natural log of e^x?

  8. myininaya
    • 5 years ago
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    int(e^x-3) =int(e^x)-int(3) =e^x-3x+C C is a constant

  9. anonymous
    • 5 years ago
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    hey could this be the final answer? u didnt use reimanns sum though

  10. myininaya
    • 5 years ago
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    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

  11. myininaya
    • 5 years ago
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    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

  12. myininaya
    • 5 years ago
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    oops mistake =(e^3-3(3))-(e^1-3(1))

  13. myininaya
    • 5 years ago
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    i have to go good luck

  14. anonymous
    • 5 years ago
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    thanks a lot!

  15. anonymous
    • 5 years ago
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    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 \[e^3-e-6\]

  16. anonymous
    • 5 years ago
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    could you please explain further?

  17. anonymous
    • 5 years ago
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    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.

  18. anonymous
    • 5 years ago
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    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. \[\int^b_a c dx=(b-a)c\]

  19. anonymous
    • 5 years ago
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    thanks a lot

  20. anonymous
    • 5 years ago
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    no problem but i still have no idea how to compute this as a reimann sum

  21. myininaya
    • 5 years ago
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    see if this helps its a geometric series

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  22. myininaya
    • 5 years ago
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    but maybe there is a mistake somewhere but i'm on the right track

  23. myininaya
    • 5 years ago
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    almost there

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  24. myininaya
    • 5 years ago
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    use l'hospital's rule and you have \[\lim_{u \rightarrow 0} \frac{ue^u}{1-e^{u}}=\lim_{u \rightarrow 0} \frac{e^u+ue^{u}}{-e^{u}}\]

  25. myininaya
    • 5 years ago
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    \[=\lim_{u \rightarrow 0} \frac{1+u}{-1}=\lim_{u \rightarrow 0} (-1-u)=-1-0=-1\]

  26. myininaya
    • 5 years ago
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    and (-1)(e-e^3)=e^3-e :) and we win!

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