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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.

  • 2 years ago
  • 2 years 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

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

    • 2 years 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

    • 2 years ago
  4. jorourk3 Group Title
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    22?

    • 2 years ago
  5. jorourk3 Group Title
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    22/3

    • 2 years ago
  6. jorourk3 Group Title
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    2/3

    • 2 years ago
  7. mark_o. Group Title
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    let me know if you have question

    • 2 years ago
  8. jorourk3 Group Title
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    it says 22/3 is wrong

    • 2 years 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=____?

    • 2 years 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

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

    • 2 years ago
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