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kjp92123

  • 5 years ago

Find the definite integral using the Fundamental Theorem of Calculus.

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  1. kjp92123
    • 5 years ago
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    \[\int\limits_{-1}^{1} e^-x (4-e^x) dx\]

  2. 92saadz
    • 5 years ago
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    see the thing is iruno how to break this ice

  3. 92saadz
    • 5 years ago
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    is exactly what i need help with

  4. kjp92123
    • 5 years ago
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    That's e^(-x) by the way And I know, I'm so lost.

  5. 92saadz
    • 5 years ago
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    haha yeh this crap makes people lost in formulas man

  6. Im_Hotep55
    • 5 years ago
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    so define the fundamental thrm of calculus; then see how that applies :)

  7. kjp92123
    • 5 years ago
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    My real question is, I don't know where to start with this problem.

  8. Im_Hotep55
    • 5 years ago
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    start by defining the FTC and see how it applies lol.... that is the start

  9. kjp92123
    • 5 years ago
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    That does nothing for me.

  10. dumbcow
    • 5 years ago
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    expand it and you will get 4e^-x - 1

  11. math93
    • 5 years ago
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    Yes, what dumbcow said. Then you can take the integral of each part.

  12. Im_Hotep55
    • 5 years ago
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    FTC simply says it CAN be done; then you apply the techniques :)

  13. kjp92123
    • 5 years ago
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    I don't know how to apply the techniques haha, that's why I'm here!

  14. Im_Hotep55
    • 5 years ago
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    the equation editor seems to have distorted the equation ; can you verify it?

  15. kjp92123
    • 5 years ago
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    I was given a take-home test, and I'm supposed to teach myself definite integrals and have it due tomorrow.

  16. dumbcow
    • 5 years ago
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    FTC says the definite integral = F(1) - F(-1) but you have to find F(x) by taking anti-derivative of f(x)

  17. kjp92123
    • 5 years ago
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    \[\int\limits_{-1}^{1} 1/(e^x) (4-e^x) dx\]

  18. math93
    • 5 years ago
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    If you integrate 4e^-x, you would get -4e^-x. Then, integrate 1 and you get x So then you have -4e^-x - x evaluated from -1 to 1

  19. Im_Hotep55
    • 5 years ago
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    \[\int\limits_{-1}^{1} \frac{1}{e^x} (4-e^x) dx\]

  20. kjp92123
    • 5 years ago
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    ^^ that

  21. Im_Hotep55
    • 5 years ago
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    frac{top}{bottom} in the editor makes for fancy fractions :)

  22. kjp92123
    • 5 years ago
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    Ooo, ok!

  23. Im_Hotep55
    • 5 years ago
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    integrate {4/(e^x) - 1} dx

  24. Im_Hotep55
    • 5 years ago
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    4 (ln(e^x)) - x

  25. dumbcow
    • 5 years ago
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    do what math93 said

  26. Im_Hotep55
    • 5 years ago
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    4x-x = 3x F(x) = 3x right?

  27. dumbcow
    • 5 years ago
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    no F(x) =-4e^-x - x

  28. Im_Hotep55
    • 5 years ago
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    close lol

  29. math93
    • 5 years ago
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    \[ \frac{-4}{e ^{x}}-x\] evaluated from -1 to 1

  30. Im_Hotep55
    • 5 years ago
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    coulda thunked that 1/u integrates to ln(u)....

  31. Im_Hotep55
    • 5 years ago
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    4e^-x is just as good i spose :)

  32. math93
    • 5 years ago
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    if you sub in your values, you get (-4e^-1 - 1)-(-4e^-1+1)

  33. Im_Hotep55
    • 5 years ago
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    i see it..... just blind in my old age

  34. kjp92123
    • 5 years ago
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    After the values are substituted in, do I just simplify?

  35. math93
    • 5 years ago
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    yes

  36. kjp92123
    • 5 years ago
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    So is the final answer (-2)?

  37. math93
    • 5 years ago
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    Yeah, that's what I got

  38. kjp92123
    • 5 years ago
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    So is the answer just (-2) by itself? Or is there anything on the opposite side of the equal sign?

  39. math93
    • 5 years ago
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    the integral of the original problem = -2, so "-2" is the final answer

  40. kjp92123
    • 5 years ago
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    Alright, I appreciate the help, I'll use this one as an example to hopefully finish the rest of the problems I have, cheers!

  41. math93
    • 5 years ago
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    Good luck!

  42. kjp92123
    • 5 years ago
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    Thank you!!

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