Find the antiderivative of f(x)= (x^3+cube root of x+3)/(3)
f'(x)= (x^3+cube root of x+3)/(3) --- forgot the prime
Distribute the 3 into both components of the numerator and integrate the two parts separately like so: \[F(x)=\int\limits_{}^{}(x^3/3)dx+\int\limits_{}^{}(\sqrt{x+3}/3)dx\] The first integral is fairly straightforward. To do the second one, you have to choose a substitution u = x + 3, find that du = dx and make the substitution to find the antiderivative of sqrt(u)/3. In the end, you should get: \[F(x)=x^4/12+2/9(x+3)^{3/2}+C\]
it says math processing error so i cant see your work
Okay, I tried doing it with a square root anyway, so it is incorrect. I'll work on it.
thank you. Im workin on it too
If f'(x) = (x^3 + cuberoot(x + 3))/3, then... f(x) = integral(x^3 / 3)dx + integral(cuberoot(x + 3) / 3)dx Find the antiderivative of each of these. The second one you must use a u-substitution of u = x + 3. f(x) = x^4 / 12 + (x + 3)^(4/3) / 4 + C However, the problem wants the antiderivative of f, so we have to take another antiderivative. I'll put the second part of the integral back into the u-substitution to get: F(x) = integral(x^4 / 12)dx + integral(u^(4/3) / 4)du + C F(x) = x^5 / 60 + 3u^(7/3) / 28 + Cx + D I hope I did that right... :/
Ha, silly me. You don't need to do the part of finding the antiderivative of the antiderivative, I don't think.
oh yea thats right, its just one antiderivative
f(x) = [x^3 + cbrt(x+3)] --------------- 3 is this the equation?
yes
thats it
please explain an easier way of looking at antiderivatives. My teacher just wants us to do trial and error. I know all the derivative rules by the way
(1/3)x^3 + (1/3)(x+3)^(1/3) sound good?
thats F(x)?
no thats a modified version of f(x); trying to get it to look more "do-able"
oh okay
so now you take it backwards?
as much as you can :) the first term is easy... (1/3)(1/4) x^4
the second one we want to make sure we get all the parts right: (S) (1/3)(x+3)^(1/3) dx (1/3) (S) (x+3)^(1/3) dx lets try to put this into a (S) u du type format.. u = x+3 du = 1 dx which is good, it means that we aint got to add nothing fancy to it :)
(1/3) (S) u^(1/3) du (1/3) (4/3)(x+3)^(4/3) does that make sense?
what is S
lol .... the integral sign is a deformed "S" so I just use that to indicate the integral sign :) (S) = integral sign... thats all
oh ok
and it should be (3/4)(x+3)^(4/3) thats better
check to make sure that derives down to our intended form...
thats the 2nd term right?
so the first term is 1/12 (x^4)
2nd term: (1/3)(3/4) (x+3)^(4/3) right? clean it up some....
youre keeping good track of this stuff :)
i can do the chain rule to clean that up?
no chain rule needed, just squish it all together to get something resembling one term :)
what do you get?
(x+1) cbrt(x+1) ------------- is what I get for the 2nd term 4
we cant get rid of the cbrt?
nope, its stuck there :) but we were aboe to pull out a chunk of that (x+3)^4 :) namely that (x+3)^3 its fine, unless we missed something. better double check our work to make sure :)
yea let me do the derivative of it
im gonna go get a coffee while you whittle away :)
kk
this is what i got: F'(x)= x^3+1/4[(x+1)[1/3(x+1)^-2/3+(x+1)^1/3]
that doesnt seem like the original
let me do this on paper to check your work...and mine :)
k sounds good
x^4 (x+3)^(4/3) F(x) = ---- + ---------- 12 4 This is what I want to "derive" to get back to our original equation....
I get: 4x^3 4 (x+3)^(1/3) f(x) = ----- + ------------- (4)(3) (4)(3) Do you see that it works? or did I miss something?
take the derivative
its not working for me
check my work, and se if its right :)
k the first term def works
the second term would be 1/12(4) 1/3(x+3)^-2/3
so that is 1/3* (1/3(x+3)^-2/3
remember to take the (1/4) to the side and derive the rest of it, dont try to confuse this with the quotient rule
im doing chain
let me try again
1 sec
the only thing to chain would be the derivative of (x+3) which equals 1 :)
1/4*4/3(x+3)
1/4*4/3(x+3)^1/3*1
thats the chain
there you go :)
i dont see how that makes the original
cross your 4s, and whats left?
(1/4)(4/3) = (1/3)
yea but its x^3 on the bottom
(x+3)^(1/3) = cbrt(x+3) right?
the 3 isnt in the cbrt. Its not suppose to be
theres no x term on the bottom.... clean it up so you can see what your looking at :)
but f(x) is (x^3+cbrt(x)+3)/x^3
now now, I asked you if I had the equation right to begin with.... and now you wanna change it on me? .... :)
lol sorry
lol .... well it does make it easier
i wrote it wrong its my fault
integratings integrating, doesnt matter if its the right one or not, well get it done :)
f(x) = (x^3+cbrt(x)+3)/x^3 we good with this one then?
yes
just remember, you start at the beginning, and when you get to the end....stop :)
k
thats from alice in wonderland, through the looking glass :)
can we set it up like this x^3*(x)^1/3+3*x^-3
x^3 +x^(1/3) +3 f(x) = --- --------- --- x^3 x^3 x^3
f(x) = 1 + x^(-8/3) + 3x^(-3) right?
which form are you using
thats right
1/3/3=-8/3
1/3-3 i mean
the rest is cake walk :) f(x) =1 + x^(-8/3) + 3x^(-3) F(x) = [x] + [3x^(-5/3)]/5 - [3x^(-2)]/2 + C good?
let me take the derivative
second term should be (-) not (+)
why negatives, everything needs to be positive that wont make sense
for instance, take the last term: (S) 3x^(-3) dx 3 x^-2 ------ do you see why its negative now? -2 <----- makes a difference
oh ok
2nd term: (S) x^(-8/3) dx x^[(-8/3 + 3/3)] x^(-5/3) 3x^(-5/3) -------------- = ------- = --------- (-8/3 + 3/3) -5/3 -5
sounds good. lol it was so confusing im sorry
usually when I miss something on a test, its cause I forgot to keep track of me signs :)
so it the F(x)=x-1/5*(3x)^-5/3-[3x)^-2]+c
?
it looks good except for this: ....-[3x)^-2]___+c somethings missing :) can you tell me what you left out? and its prolly just a typo error...
-1/2*(3x)^-2?
thats what I was looking for :) good job, now get the derivative and see if it matches....which it will :)
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