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Math2400

  • one year ago

can someone explain this? I don't understand my instructors steps:

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  1. Math2400
    • one year ago
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    suppose that f(1) =2. f(4)=5, f'(4)=3, and f'' is continuous. Find the value of \[\int\limits_{1}^{4} xf''(x)dx\]

  2. Math2400
    • one year ago
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    the answer is 9 but im not sure how my prof. did it. i think he tried taking the anti-deriv. first. or used some kind of property. cuz he got something like: |dw:1433615204902:dw|

  3. Math2400
    • one year ago
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    but idk how he got that>< can anyone explain?

  4. Math2400
    • one year ago
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    @Luigi0210 would u be able help?

  5. Math2400
    • one year ago
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    @ganeshie8 another one if you're free :) i'm studying for my final><

  6. amistre64
    • one year ago
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    looks like a by parts application

  7. Luigi0210
    • one year ago
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    ^ -.-

  8. amistre64
    • one year ago
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    what is the derivative of a product?

  9. amistre64
    • one year ago
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    lets take the derivative of the product of 2 function, f and g [fg]' = ??

  10. amistre64
    • one year ago
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    or, to better line up with your specifics .. [gf]' = ??

  11. Math2400
    • one year ago
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    g'f+gf'

  12. amistre64
    • one year ago
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    good [gf]' = g'f + gf' now integrate both sides

  13. amistre64
    • one year ago
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    [gf]' integrates back in gf and we have the sum of integration on the right \[gf=\int g'f~+~\int gf'\] now, we want to know how to take the integration of gf', so lets subtract to get it all by itself ... and we end up with the by parts formulation \[\int gf'=gf-\int g'f\]

  14. amistre64
    • one year ago
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    let g=x, and f' = f''

  15. amistre64
    • one year ago
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    the letters we named our function are immaterial, one usual notation is to use u and v \[\int uv'=uv-\int vu'\]

  16. ybarrap
    • one year ago
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    I think that you will also need \(f'(1)=0\) as a \(\bf{given}\) later in this process.

  17. phi
    • one year ago
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    Hopefully you have leaned integration by parts (which is what Amistre was deriving) https://en.wikipedia.org/wiki/Integration_by_parts You use that "rule" to get your result

  18. phi
    • one year ago
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    you also need to now that \[ \int \frac{d^2 f(x)}{dx^2} \ dx = \frac{d f(x)}{dx} = f'(x)\]

  19. phi
    • one year ago
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    *know

  20. ybarrap
    • one year ago
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    Applying the rules of @amistre64 we have $$ [xf']'=x'f'+xf''=f'+xf'' $$ Then $$ \int [xf']'dx=xf'\\ =\int \left (x'f'+xf''\right ) dx\\ =\int \left (f'+xf''\right ) dx\\ $$ From which you get what you have written above. Now evaluate. You are missing an initial condition: \(f'(1)=0\). Using this, you'll get 9 after evaluation.

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