Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.) f(x) = x(6 − x)2 for my answer I got X^4/4-4x^3+18x^2+C is this right?

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Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.) f(x) = x(6 − x)2 for my answer I got X^4/4-4x^3+18x^2+C is this right?

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Did you find the derivative to check yourself?
To keep things simple, you may try u-substitution \(u = 6-x\)
no i didn't find the derivative to check my answer, i would find the derivative of my problem right?

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yeah you can differentiate your answer and see if it is x^3-12x^2+36x which is the form I think you put it in to integrate (by the way if you have heard of substitutions to make integrals like this nicer looking (or I mean easier) at ganeshie8's note )
nvm if you haven't heard of u-substition before your work looks good, i presume you have used the formula for antiderivative of \(x^n\) to double check, differentiate the answer as freckles said
he might heard of it I don't know
Im not sure i have, but i took the derivative and got x^2-12x+36x :)
x^3-12x^2+36x?
yeah its just that i had x(x^2-12x+36)
and yes I already checked your answer integrating thingy but your problem said to also check your solution by differentiating so now you done both things your problem asked for
got it right! thanks!!
how do you know ` x(x^2-12x+36) ` is same as the original function ?
that might be a dumb q hmm
x(6-x)^2 think he expanded the square thingy first
she* and yeah i did
Ahh looks great she :D
oops sorry
do you want to see the sub thingy mentioned by ganeshie8 for fun?
Its okay! sure is it easier than finding the derivative?
\[\int\limits x (6-x)^2 dx \\ \text{ Let } u=6-x \\ \frac{du}{dx}=0-1 \\ \frac{du}{dx}=-1 \\ du=-1 dx \\ \text{ multiply both sides by -1 } \\ -du=dx \\ \text{ now recall if } u=6-x \text{ then } x=6-u \\ \text{ so we have } \\ \int\limits x (6-x)^2 dx=\int\limits (6-u)u^2 (-du) \\ =\int\limits (u-6)u^2 du\] so now you have less multiplication to do
\[=\int\limits (u-6)u^2 du =\int\limits (u^3-6u) du=\frac{u^4}{4}-3u^2+C \\ \text{ now remember } u=6-x \\ \text{ so you have } \\ =\frac{(6-x)^4}{4}-3(6-x)^2+C\] and it is less multiplication if you get to leave your answer like this :)
\[ x(6 − x)^2 = x(x^2-12x+36) = x^3-12x^2+36x \]Then use power rule. No tricks needed. Your anti-derivative looks correct.
Oh okay, I didnt understand the substitution part at first, but i get it now! thanks for explaining!
oops I didn't multiply correclty
my answer is off because when I did -6(u^2) I put -6u and so I integrated the wrong thingy
but that is sorta how works above if you can multiply correctly :p
you should ask why on earth substitution is any better than your original method @gaba
try this if you're loving antiderivatives : \[\int x(6-x)^{9999999}\,dx = ?\]
I'm deff not loving them lol so Im working on this right now, Find the most general antiderivative of the function 2rootx+6cosx I got 4x^1/2+6-sinx
im not sure i did it right though
differentiate your answer and see if you get back the original function
also what happened to the constant, C
nvm I just did it again and got 4x^3/2/3 +6 -sinx +C
you're doing calculus, that means you're not in highschool anymore time to use proper notation
4x^ `3/2/3` +6 -sinx +C what does that even mean
[(4x^3/2)/3] +6-sinx+C Btw you can take calculus in high school, just saying
Much better, but still it is not mathematically correct
pretty sure you meant [(4x^3/2)/3] +6(-sinx)+C
differentiate that and see if you get back the original function (there is a mistake, so you wont get back the original function)
why is it +6(sin(x)) and not -sin(x)?
\[f(x)=2 \sqrt{x}+6 \cos(x) \\ \text{ or we can write } \\ f(x)=2 x^\frac{1}{2}+6 \cos(x)\] is this what you are playing with?
yes!
anyways ( ? )'=cos(x) the ?=sin(x) right? (sin(x))'=cos(x) so the antiderivative of cos(x) is sin(x)+C if I had ( ? )'=sin(x) the ?=-cos(x) (-cos(x))'=sin(x) so the antiderivative of sin(x) is -cos(x)+C now if you had ( ? )'=-cos(x) then ?=-sin(x) (-sin(x))'=-cos(x) so the antiderivative of -cos(x) is -sin(x)+C anyways you had find the antiderivative of 2x^(1/2) which looks like you were going for: 4x^(3/2)/3 +k but to find the antiderivative of 6cos(x) you need to recall what can you take derivative of that will give you 6cos(x) you can bring down the 6( and put the antiderivative of cos(x)) here )
forgot to put + some constant at the end there
oh okay, is that to make sure I had it right I checked on mathway for the derivative of cos(x) and it said it was -sin(x) (which is what I thought it was before checking so it made sense) but i typed in +sin(x) and got it right, so thanks for explaining everything!
oh so you got derivative and antiderivative mixed up
but just so you know if it was taking derivative of 6cos(x) you would put -6sin(x) and not 6-sin(x)
but yeah antiderivative of 6cos(x) would be 6sin(x) since the derivative of 6sin(x) is 6cos(x)
the antiderivative of 6cos(x) would be 6sin(x)+C since the derivative of 6sin(x)+C is 6cos(x) *
the most general antiderivative *
Okay I understand now! thanks a lot!!
np

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