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anonymous

  • 4 years ago

Can some please help me with this: Suppose f(3)=4, f(8)=8, f'(3)=2, f'(8)=9, and f'' is continuous. Find the value of the definite integral

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{3}^{8}\] xf''(x)

  2. anonymous
    • 4 years ago
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    I need this broken down in layman terms

  3. amistre64
    • 4 years ago
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    well, if we try to use this according to the definition of integration .. what can we get?

  4. amistre64
    • 4 years ago
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    \[f(x)=\int f'(x)dx\]is the basic understanding of an integral

  5. anonymous
    • 4 years ago
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    I don't understand what to do with the function values

  6. amistre64
    • 4 years ago
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    it would appear to me that the 3 to 8 is the limits of integration

  7. amistre64
    • 4 years ago
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    \[\int_{3}^{8}x\ f''(x) dx\] is what we have to work with and all we have to determine is what to do with the "x" part

  8. amistre64
    • 4 years ago
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    i might have to logout and back in to do this ...

  9. anonymous
    • 4 years ago
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    ok

  10. amistre64
    • 4 years ago
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    the math still aint processing from the latex on my end; you see normal equations on your end?

  11. anonymous
    • 4 years ago
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    yes I can see them.

  12. amistre64
    • 4 years ago
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    good :) well then, lets see if we can move ahead with this then i think we can apply integration by parts to this .. just a gut felling

  13. amistre64
    • 4 years ago
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    \[\int u dv =uv-\int v du\]

  14. anonymous
    • 4 years ago
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    ok so would i just use xf''(x) as my function to integrate?

  15. amistre64
    • 4 years ago
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    to wit: u = x v = f'(x) du = dx dv = f''(x) dx yes

  16. anonymous
    • 4 years ago
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    so when I integrate that is where the values that were given will be used?

  17. amistre64
    • 4 years ago
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    \[\int xf''(x)=xf'(x)-\int f'(x)dx\] \[\int xf''(x)=xf'(x)-f(x)\]

  18. amistre64
    • 4 years ago
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    yes

  19. amistre64
    • 4 years ago
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    im sure that looks alot more inspiring on your end lol

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  20. anonymous
    • 4 years ago
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    I shows up normal on my end

  21. amistre64
    • 4 years ago
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    (8 f'(8)) - f(8) ) - (3 f'(3)-f(3)) should be our results

  22. anonymous
    • 4 years ago
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    ok I didn't get that far but how did you get rid of the x?

  23. amistre64
    • 4 years ago
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    I didnt; i just used it in the integration by parts formula. there is no reason to get rid of it since the interval speaks in x to begin with

  24. anonymous
    • 4 years ago
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    ok i see know you used the limit that you are evaluating the integral?

  25. amistre64
    • 4 years ago
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    yep

  26. anonymous
    • 4 years ago
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    I got 53, is that correct?

  27. anonymous
    • 4 years ago
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    I see my error, I am about to recalculate it

  28. amistre64
    • 4 years ago
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    8*9 - 8 - (3*2-4) 72-8 - (6-4) 64 - 2 = 62

  29. amistre64
    • 4 years ago
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    i hope that makes more sense, I have to be heading out now. good luck :)

  30. anonymous
    • 4 years ago
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    I accidently used 8*8-9-)-(3*2-4). Thank you that was really helpful.

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