anonymous
  • anonymous
Find the antiderivative of f(x)= (x^3+cube root of x+3)/(3)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
f'(x)= (x^3+cube root of x+3)/(3) --- forgot the prime
anonymous
  • anonymous
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: \[F(x)=x^4/12+2/9(x+3)^{3/2}+C\]
anonymous
  • anonymous
it says math processing error so i cant see your work

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anonymous
  • anonymous
Okay, I tried doing it with a square root anyway, so it is incorrect. I'll work on it.
anonymous
  • anonymous
thank you. Im workin on it too
anonymous
  • anonymous
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... :/
anonymous
  • anonymous
Ha, silly me. You don't need to do the part of finding the antiderivative of the antiderivative, I don't think.
anonymous
  • anonymous
oh yea thats right, its just one antiderivative
amistre64
  • amistre64
f(x) = [x^3 + cbrt(x+3)] --------------- 3 is this the equation?
anonymous
  • anonymous
yes
anonymous
  • anonymous
thats it
anonymous
  • anonymous
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
amistre64
  • amistre64
(1/3)x^3 + (1/3)(x+3)^(1/3) sound good?
anonymous
  • anonymous
thats F(x)?
amistre64
  • amistre64
no thats a modified version of f(x); trying to get it to look more "do-able"
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
so now you take it backwards?
amistre64
  • amistre64
as much as you can :) the first term is easy... (1/3)(1/4) x^4
amistre64
  • amistre64
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 :)
amistre64
  • amistre64
(1/3) (S) u^(1/3) du (1/3) (4/3)(x+3)^(4/3) does that make sense?
anonymous
  • anonymous
what is S
amistre64
  • amistre64
lol .... the integral sign is a deformed "S" so I just use that to indicate the integral sign :) (S) = integral sign... thats all
anonymous
  • anonymous
oh ok
amistre64
  • amistre64
and it should be (3/4)(x+3)^(4/3) thats better
amistre64
  • amistre64
check to make sure that derives down to our intended form...
anonymous
  • anonymous
thats the 2nd term right?
anonymous
  • anonymous
so the first term is 1/12 (x^4)
amistre64
  • amistre64
2nd term: (1/3)(3/4) (x+3)^(4/3) right? clean it up some....
amistre64
  • amistre64
youre keeping good track of this stuff :)
anonymous
  • anonymous
i can do the chain rule to clean that up?
amistre64
  • amistre64
no chain rule needed, just squish it all together to get something resembling one term :)
anonymous
  • anonymous
what do you get?
amistre64
  • amistre64
(x+1) cbrt(x+1) ------------- is what I get for the 2nd term 4
anonymous
  • anonymous
we cant get rid of the cbrt?
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
yea let me do the derivative of it
amistre64
  • amistre64
im gonna go get a coffee while you whittle away :)
anonymous
  • anonymous
kk
anonymous
  • anonymous
this is what i got: F'(x)= x^3+1/4[(x+1)[1/3(x+1)^-2/3+(x+1)^1/3]
anonymous
  • anonymous
that doesnt seem like the original
amistre64
  • amistre64
let me do this on paper to check your work...and mine :)
anonymous
  • anonymous
k sounds good
amistre64
  • amistre64
x^4 (x+3)^(4/3) F(x) = ---- + ---------- 12 4 This is what I want to "derive" to get back to our original equation....
amistre64
  • amistre64
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?
anonymous
  • anonymous
take the derivative
anonymous
  • anonymous
its not working for me
amistre64
  • amistre64
check my work, and se if its right :)
anonymous
  • anonymous
k the first term def works
anonymous
  • anonymous
the second term would be 1/12(4) 1/3(x+3)^-2/3
anonymous
  • anonymous
so that is 1/3* (1/3(x+3)^-2/3
amistre64
  • amistre64
remember to take the (1/4) to the side and derive the rest of it, dont try to confuse this with the quotient rule
anonymous
  • anonymous
im doing chain
anonymous
  • anonymous
let me try again
anonymous
  • anonymous
1 sec
amistre64
  • amistre64
the only thing to chain would be the derivative of (x+3) which equals 1 :)
anonymous
  • anonymous
1/4*4/3(x+3)
anonymous
  • anonymous
1/4*4/3(x+3)^1/3*1
anonymous
  • anonymous
thats the chain
amistre64
  • amistre64
there you go :)
anonymous
  • anonymous
i dont see how that makes the original
amistre64
  • amistre64
cross your 4s, and whats left?
amistre64
  • amistre64
(1/4)(4/3) = (1/3)
anonymous
  • anonymous
yea but its x^3 on the bottom
amistre64
  • amistre64
(x+3)^(1/3) = cbrt(x+3) right?
anonymous
  • anonymous
the 3 isnt in the cbrt. Its not suppose to be
amistre64
  • amistre64
theres no x term on the bottom.... clean it up so you can see what your looking at :)
anonymous
  • anonymous
but f(x) is (x^3+cbrt(x)+3)/x^3
amistre64
  • amistre64
now now, I asked you if I had the equation right to begin with.... and now you wanna change it on me? .... :)
anonymous
  • anonymous
lol sorry
amistre64
  • amistre64
lol .... well it does make it easier
anonymous
  • anonymous
i wrote it wrong its my fault
amistre64
  • amistre64
integratings integrating, doesnt matter if its the right one or not, well get it done :)
amistre64
  • amistre64
f(x) = (x^3+cbrt(x)+3)/x^3 we good with this one then?
anonymous
  • anonymous
yes
amistre64
  • amistre64
just remember, you start at the beginning, and when you get to the end....stop :)
anonymous
  • anonymous
k
amistre64
  • amistre64
thats from alice in wonderland, through the looking glass :)
anonymous
  • anonymous
can we set it up like this x^3*(x)^1/3+3*x^-3
amistre64
  • amistre64
x^3 +x^(1/3) +3 f(x) = --- --------- --- x^3 x^3 x^3
amistre64
  • amistre64
f(x) = 1 + x^(-8/3) + 3x^(-3) right?
anonymous
  • anonymous
which form are you using
anonymous
  • anonymous
thats right
anonymous
  • anonymous
1/3/3=-8/3
anonymous
  • anonymous
1/3-3 i mean
amistre64
  • amistre64
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?
anonymous
  • anonymous
let me take the derivative
amistre64
  • amistre64
second term should be (-) not (+)
anonymous
  • anonymous
why negatives, everything needs to be positive that wont make sense
amistre64
  • amistre64
for instance, take the last term: (S) 3x^(-3) dx 3 x^-2 ------ do you see why its negative now? -2 <----- makes a difference
anonymous
  • anonymous
oh ok
amistre64
  • amistre64
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
anonymous
  • anonymous
sounds good. lol it was so confusing im sorry
amistre64
  • amistre64
usually when I miss something on a test, its cause I forgot to keep track of me signs :)
anonymous
  • anonymous
so it the F(x)=x-1/5*(3x)^-5/3-[3x)^-2]+c
anonymous
  • anonymous
?
amistre64
  • amistre64
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...
anonymous
  • anonymous
-1/2*(3x)^-2?
amistre64
  • amistre64
thats what I was looking for :) good job, now get the derivative and see if it matches....which it will :)
anonymous
  • anonymous
lol my brain isnt working forwards now
amistre64
  • amistre64
F(x) = x - [3x^(-5/3)]/5 - [3x^-2]/2 + C f(x) = 1 - [(3/5)(-5/3) x^(-8/3)] - [(3/2)(-2) x^-3 ] f(x) = 1 + x^(-8/3) +3x^(-3) which we can modify back to your original stuff....
anonymous
  • anonymous
oh ok
amistre64
  • amistre64
does it make sense to you? or are you just agreeing with me :)
anonymous
  • anonymous
how does x^(-8/3) become cbrt of x?
anonymous
  • anonymous
never mind, you divide everything by x^3
anonymous
  • anonymous
correct?
amistre64
  • amistre64
put all your negative exponents underneath again.... but make sure you take out the x^(-3) from x^(-8/3) remember when bases multiply together, their exponents "add" up. to split them simply undo the process like this: x^(-8/3) = x^(1/3) x^(-9/3)
amistre64
  • amistre64
yes, in essense, yes :)
anonymous
  • anonymous
how does the 1 become x^3
anonymous
  • anonymous
1/x^3=x^3 with the common denomonator
anonymous
  • anonymous
omg i am acting dumb, sorry
amistre64
  • amistre64
(x^3)/(x^3) = 1 right? thats where we got it from
anonymous
  • anonymous
yea thats right 1/x^3= x^3+1/x^3
amistre64
  • amistre64
had to go hunt down which version we were using :)
anonymous
  • anonymous
yea it works out, complicated but works out
amistre64
  • amistre64
its easier with a fresh pair of eyes :)
anonymous
  • anonymous
yea thank you sooooooo much
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
Have a good night, ill ttyl

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