Question

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

Answer 1

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

Answer 2

what do you mean by integrating them?

Answer 3

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

Answer 4

22?

Answer 5

22/3

Answer 6

2/3

Answer 7

let me know if you have question

Answer 8

it says 22/3 is wrong

Answer 9

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=____?

Answer 10

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

Answer 11

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