Find the antiderivative of
f(x)= (x^3+cube root of x+3)/(3)

- anonymous

Find the antiderivative of
f(x)= (x^3+cube root of x+3)/(3)

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- anonymous

f'(x)= (x^3+cube root of x+3)/(3) --- forgot the prime

- 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

it says math processing error so i cant see your work

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- anonymous

Okay, I tried doing it with a square root anyway, so it is incorrect. I'll work on it.

- anonymous

thank you. Im workin on it too

- 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

Ha, silly me. You don't need to do the part of finding the antiderivative of the antiderivative, I don't think.

- anonymous

oh yea thats right, its just one antiderivative

- amistre64

f(x) = [x^3 + cbrt(x+3)]
---------------
3
is this the equation?

- anonymous

yes

- anonymous

thats it

- 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

(1/3)x^3 + (1/3)(x+3)^(1/3) sound good?

- anonymous

thats F(x)?

- amistre64

no thats a modified version of f(x); trying to get it to look more "do-able"

- anonymous

oh okay

- anonymous

so now you take it backwards?

- amistre64

as much as you can :)
the first term is easy...
(1/3)(1/4) x^4

- 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

(1/3) (S) u^(1/3) du
(1/3) (4/3)(x+3)^(4/3) does that make sense?

- anonymous

what is S

- 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

oh ok

- amistre64

and it should be (3/4)(x+3)^(4/3) thats better

- amistre64

check to make sure that derives down to our intended form...

- anonymous

thats the 2nd term right?

- anonymous

so the first term is 1/12 (x^4)

- amistre64

2nd term:
(1/3)(3/4) (x+3)^(4/3) right? clean it up some....

- amistre64

youre keeping good track of this stuff :)

- anonymous

i can do the chain rule to clean that up?

- amistre64

no chain rule needed, just squish it all together to get something resembling one term :)

- anonymous

what do you get?

- amistre64

(x+1) cbrt(x+1)
------------- is what I get for the 2nd term
4

- anonymous

we cant get rid of the cbrt?

- 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

yea let me do the derivative of it

- amistre64

im gonna go get a coffee while you whittle away :)

- anonymous

kk

- 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

that doesnt seem like the original

- amistre64

let me do this on paper to check your work...and mine :)

- anonymous

k sounds good

- 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

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

take the derivative

- anonymous

its not working for me

- amistre64

check my work, and se if its right :)

- anonymous

k the first term def works

- anonymous

the second term would be 1/12(4) 1/3(x+3)^-2/3

- anonymous

so that is 1/3* (1/3(x+3)^-2/3

- 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

im doing chain

- anonymous

let me try again

- anonymous

1 sec

- amistre64

the only thing to chain would be the derivative of (x+3) which equals 1 :)

- anonymous

1/4*4/3(x+3)

- anonymous

1/4*4/3(x+3)^1/3*1

- anonymous

thats the chain

- amistre64

there you go :)

- anonymous

i dont see how that makes the original

- amistre64

cross your 4s, and whats left?

- amistre64

(1/4)(4/3) = (1/3)

- anonymous

yea but its x^3 on the bottom

- amistre64

(x+3)^(1/3) = cbrt(x+3) right?

- anonymous

the 3 isnt in the cbrt. Its not suppose to be

- amistre64

theres no x term on the bottom.... clean it up so you can see what your looking at :)

- anonymous

but f(x) is (x^3+cbrt(x)+3)/x^3

- amistre64

now now, I asked you if I had the equation right to begin with.... and now you wanna change it on me? .... :)

- anonymous

lol sorry

- amistre64

lol .... well it does make it easier

- anonymous

i wrote it wrong its my fault

- amistre64

integratings integrating, doesnt matter if its the right one or not, well get it done :)

- amistre64

f(x) = (x^3+cbrt(x)+3)/x^3 we good with this one then?

- anonymous

yes

- amistre64

just remember, you start at the beginning, and when you get to the end....stop :)

- anonymous

k

- amistre64

thats from alice in wonderland, through the looking glass :)

- anonymous

can we set it up like this x^3*(x)^1/3+3*x^-3

- amistre64

x^3 +x^(1/3) +3
f(x) = --- --------- ---
x^3 x^3 x^3

- amistre64

f(x) = 1 + x^(-8/3) + 3x^(-3) right?

- anonymous

which form are you using

- anonymous

thats right

- anonymous

1/3/3=-8/3

- anonymous

1/3-3 i mean

- 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

let me take the derivative

- amistre64

second term should be (-) not (+)

- anonymous

why negatives, everything needs to be positive that wont make sense

- 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

oh ok

- 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

sounds good. lol it was so confusing im sorry

- amistre64

usually when I miss something on a test, its cause I forgot to keep track of me signs :)

- anonymous

so it the F(x)=x-1/5*(3x)^-5/3-[3x)^-2]+c

- anonymous

?

- 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

-1/2*(3x)^-2?

- amistre64

thats what I was looking for :) good job, now get the derivative and see if it matches....which it will :)

- anonymous

lol my brain isnt working forwards now

- 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

oh ok

- amistre64

does it make sense to you? or are you just agreeing with me :)

- anonymous

how does x^(-8/3) become cbrt of x?

- anonymous

never mind, you divide everything by x^3

- anonymous

correct?

- 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

yes, in essense, yes :)

- anonymous

how does the 1 become x^3

- anonymous

1/x^3=x^3 with the common denomonator

- anonymous

omg i am acting dumb, sorry

- amistre64

(x^3)/(x^3) = 1 right? thats where we got it from

- anonymous

yea thats right 1/x^3= x^3+1/x^3

- amistre64

had to go hunt down which version we were using :)

- anonymous

yea it works out, complicated but works out

- amistre64

its easier with a fresh pair of eyes :)

- anonymous

yea thank you sooooooo much

- amistre64

youre welcome :)

- anonymous

Have a good night, ill ttyl

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