Question

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

Answer 1

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

Answer 2

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

Answer 3

it says math processing error so i cant see your work

Answer 4

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

Answer 5

thank you. Im workin on it too

Answer 6

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... :/

Answer 7

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

Answer 8

oh yea thats right, its just one antiderivative

Answer 9

f(x) = [x^3 + cbrt(x+3)]
---------------
3

is this the equation?

Answer 10

yes

Answer 11

thats it

Answer 12

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

Answer 13

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

Answer 14

thats F(x)?

Answer 15

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

Answer 16

oh okay

Answer 17

so now you take it backwards?

Answer 18

as much as you can :)

the first term is easy...

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

Answer 19

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 :)

Answer 20

(1/3) (S) u^(1/3) du

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

Answer 21

what is S

Answer 22

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

(S) = integral sign... thats all

Answer 23

oh ok

Answer 24

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

Answer 25

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

Answer 26

thats the 2nd term right?

Answer 27

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

Answer 28

2nd term:

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

Answer 29

youre keeping good track of this stuff :)

Answer 30

i can do the chain rule to clean that up?

Answer 31

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

Answer 32

what do you get?

Answer 33

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

Answer 34

we cant get rid of the cbrt?

Answer 35

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 :)

Answer 36

yea let me do the derivative of it

Answer 37

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

Answer 38

kk

Answer 39

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

Answer 40

that doesnt seem like the original

Answer 41

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

Answer 42

k sounds good

Answer 43

x^4 (x+3)^(4/3)
F(x) = ---- + ----------
12 4

This is what I want to "derive" to get back to our original equation....

Answer 44

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?

Answer 45

take the derivative

Answer 46

its not working for me

Answer 47

check my work, and se if its right :)

Answer 48

k the first term def works

Answer 49

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

Answer 50

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

Answer 51

remember to take the (1/4) to the side and derive the rest of it, dont try to confuse this with the quotient rule

Answer 52

im doing chain

Answer 53

let me try again

Answer 54

1 sec

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

thats the chain

Answer 59

there you go :)

Answer 60

i dont see how that makes the original

Answer 61

cross your 4s, and whats left?

Answer 62

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

Answer 63

yea but its x^3 on the bottom

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

lol sorry

Answer 70

lol .... well it does make it easier

Answer 71

i wrote it wrong its my fault

Answer 72

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

Answer 73

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

Answer 74

yes

Answer 75

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

Answer 76

k

Answer 77

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

Answer 78

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

Answer 79

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

Answer 80

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

Answer 81

which form are you using

Answer 82

thats right

Answer 83

1/3/3=-8/3

Answer 84

1/3-3 i mean

Answer 85

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?

Answer 86

let me take the derivative

Answer 87

second term should be (-) not (+)

Answer 88

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

Answer 89

for instance, take the last term:

(S) 3x^(-3) dx

3 x^-2
------ do you see why its negative now?
-2 <----- makes a difference

Answer 90

oh ok

Answer 91

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

Answer 92

sounds good. lol it was so confusing im sorry

Answer 93

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

Answer 94

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

Answer 95

?

Answer 96

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...

Answer 97

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

Answer 98

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