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## anonymous 4 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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1. anonymous

ok , try integrating them, then and set them equal to the right hand side let see what happen

2. anonymous

what do you mean by integrating them?

3. anonymous

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

4. anonymous

22?

5. anonymous

22/3

6. anonymous

2/3

7. anonymous

let me know if you have question

8. anonymous

it says 22/3 is wrong

9. anonymous

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=____?

10. anonymous

im sorry, on note , that was du instead of dy. therefore dU=11dx and 1/11 is to balanced it from the original problem

11. anonymous

@ jorourk3 let me know if you have question,.... :D

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