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jorourk3 Group Title

antiderivative of (11x+17)^2 = k(11x+17)^3 + C Find the exact value of k that makes the antidifferentiation formula true.

  • one year ago
  • one year ago

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  1. mark_o. Group Title
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    ok , try integrating them, then and set them equal to the right hand side let see what happen

    • one year ago
  2. jorourk3 Group Title
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    what do you mean by integrating them?

    • one year ago
  3. jorourk3 Group Title
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    im sorry but i dont understand what you are talking about...we just learned this stuff today and i never really got the du dx references...i just solve differentation by using the rules we are given ive been solving these equations (solve for k) but finding the antiderivative of the left side and whatever constant i got i plugged into k on the right .....but for this problem the constant i get is 11/3 which does not give you the left side when plugged into k and differentiated

    • one year ago
  4. jorourk3 Group Title
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    22?

    • one year ago
  5. jorourk3 Group Title
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    22/3

    • one year ago
  6. jorourk3 Group Title
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    2/3

    • one year ago
  7. mark_o. Group Title
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    let me know if you have question

    • one year ago
  8. jorourk3 Group Title
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    it says 22/3 is wrong

    • one year ago
  9. mark_o. Group Title
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    integral (11x+17)^2 dx \[\int\limits_{}^{}(11x+17)^{2}dx\] let u=(11x+17) and du= 11 dx so that \[\int\limits_{}^{}u ^{2}du= \frac{ u ^{3} }{ 3 }+C\] since we can sub u and du we have \[\frac{ 1 }{ 11 } \int\limits_{}^{}(11x+17)^{2}[11dx]\] note dy=11dx and 1/11 is to balanced it from the original problem therefore we have \[\frac{ 1 }{ 11 }\int\limits_{}^{}u ^{2}du=\frac{ 1 }{ 11 }\frac{ u ^{3} }{ 3 }+C\] \[=\frac{ 1 }{ 33 }u ^{3}+C\] now sub u=(11x+17) we get \[=\frac{ 1 }{ 33 }(11x+17)^{3}+C=K(11x+17)^{3}+c\] can you notice K=____?

    • one year ago
  10. mark_o. Group Title
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    im sorry, on note , that was du instead of dy. therefore dU=11dx and 1/11 is to balanced it from the original problem

    • one year ago
  11. mark_o. Group Title
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    @ jorourk3 let me know if you have question,.... :D

    • one year ago
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