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jorourk3

  • 2 years ago

antiderivative(18)/(5-x) = kln[5-x]+c Find the exact value of k that makes the antidifferentiation formula true.

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  1. jorourk3
    • 2 years ago
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    i got 18 but is wrong

  2. mark_o.
    • 2 years ago
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    @ jorourk3 \[\int\limits_{}^{}\frac{ 18 }{ 5-x }=-18\ln(x-5)+C\]=kln[5-x]+c seems like k=-1 ? yes or no?

  3. mark_o.
    • 2 years ago
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    im sorry its -18 :D

  4. jorourk3
    • 2 years ago
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    k would be -18 then by what your saying...wouldnt it? that is also absolute value signs around ln[x-5] btw...does that change it from what you are saying?

  5. jorourk3
    • 2 years ago
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    its correct! thanks! i dont understand why it is negative though?

  6. mark_o.
    • 2 years ago
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    its negative bec you let u=5-x then du= -dx

  7. mark_o.
    • 2 years ago
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    or du/dx=-1

  8. jorourk3
    • 2 years ago
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    ok thanks....can you help me with a similar one please? :)

  9. mark_o.
    • 2 years ago
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    ok yw ..:D just post them again so that others can see them

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