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Mathematics 51 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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: \[F(x)=x^4/12+2/9(x+3)^{3/2}+C\]

OpenStudy (anonymous):

it says math processing error so i cant see your work

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thank you. Im workin on it too

OpenStudy (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... :/

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh yea thats right, its just one antiderivative

OpenStudy (amistre64):

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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

thats it

OpenStudy (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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

thats F(x)?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

oh okay

OpenStudy (anonymous):

so now you take it backwards?

OpenStudy (amistre64):

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

OpenStudy (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 :)

OpenStudy (amistre64):

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

OpenStudy (anonymous):

what is S

OpenStudy (amistre64):

lol .... the integral sign is a deformed "S" so I just use that to indicate the integral sign :) (S) = integral sign... thats all

OpenStudy (anonymous):

oh ok

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

thats the 2nd term right?

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

youre keeping good track of this stuff :)

OpenStudy (anonymous):

i can do the chain rule to clean that up?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

what do you get?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

we cant get rid of the cbrt?

OpenStudy (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 :)

OpenStudy (anonymous):

yea let me do the derivative of it

OpenStudy (amistre64):

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

OpenStudy (anonymous):

kk

OpenStudy (anonymous):

this is what i got: F'(x)= x^3+1/4[(x+1)[1/3(x+1)^-2/3+(x+1)^1/3]

OpenStudy (anonymous):

that doesnt seem like the original

OpenStudy (amistre64):

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

OpenStudy (anonymous):

k sounds good

OpenStudy (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....

OpenStudy (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?

OpenStudy (anonymous):

take the derivative

OpenStudy (anonymous):

its not working for me

OpenStudy (amistre64):

check my work, and se if its right :)

OpenStudy (anonymous):

k the first term def works

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

im doing chain

OpenStudy (anonymous):

let me try again

OpenStudy (anonymous):

1 sec

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thats the chain

OpenStudy (amistre64):

there you go :)

OpenStudy (anonymous):

i dont see how that makes the original

OpenStudy (amistre64):

cross your 4s, and whats left?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

yea but its x^3 on the bottom

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

lol sorry

OpenStudy (amistre64):

lol .... well it does make it easier

OpenStudy (anonymous):

i wrote it wrong its my fault

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

yes

OpenStudy (amistre64):

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

OpenStudy (anonymous):

k

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

which form are you using

OpenStudy (anonymous):

thats right

OpenStudy (anonymous):

1/3/3=-8/3

OpenStudy (anonymous):

1/3-3 i mean

OpenStudy (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?

OpenStudy (anonymous):

let me take the derivative

OpenStudy (amistre64):

second term should be (-) not (+)

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

oh ok

OpenStudy (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

OpenStudy (anonymous):

sounds good. lol it was so confusing im sorry

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

?

OpenStudy (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...

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

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