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

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mark_o.
 2 years ago
Best ResponseYou've already chosen the best response.0ok , try integrating them, then and set them equal to the right hand side let see what happen

jorourk3
 2 years ago
Best ResponseYou've already chosen the best response.0what do you mean by integrating them?

jorourk3
 2 years ago
Best ResponseYou've already chosen the best response.0im 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

mark_o.
 2 years ago
Best ResponseYou've already chosen the best response.0let me know if you have question

mark_o.
 2 years ago
Best ResponseYou've already chosen the best response.0integral (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=____?

mark_o.
 2 years ago
Best ResponseYou've already chosen the best response.0im sorry, on note , that was du instead of dy. therefore dU=11dx and 1/11 is to balanced it from the original problem

mark_o.
 2 years ago
Best ResponseYou've already chosen the best response.0@ jorourk3 let me know if you have question,.... :D
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