## anonymous 5 years ago Find the antiderivative of f(x)= (x^3+cube root of x+3)/(3)

1. anonymous

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

2. 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: $F(x)=x^4/12+2/9(x+3)^{3/2}+C$

3. anonymous

it says math processing error so i cant see your work

4. anonymous

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

5. anonymous

thank you. Im workin on it too

6. 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... :/

7. anonymous

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

8. anonymous

oh yea thats right, its just one antiderivative

9. amistre64

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

10. anonymous

yes

11. anonymous

thats it

12. 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

13. amistre64

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

14. anonymous

thats F(x)?

15. amistre64

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

16. anonymous

oh okay

17. anonymous

so now you take it backwards?

18. amistre64

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

19. 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 :)

20. amistre64

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

21. anonymous

what is S

22. amistre64

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

23. anonymous

oh ok

24. amistre64

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

25. amistre64

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

26. anonymous

thats the 2nd term right?

27. anonymous

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

28. amistre64

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

29. amistre64

youre keeping good track of this stuff :)

30. anonymous

i can do the chain rule to clean that up?

31. amistre64

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

32. anonymous

what do you get?

33. amistre64

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

34. anonymous

we cant get rid of the cbrt?

35. 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 :)

36. anonymous

yea let me do the derivative of it

37. amistre64

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

38. anonymous

kk

39. anonymous

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

40. anonymous

that doesnt seem like the original

41. amistre64

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

42. anonymous

k sounds good

43. 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....

44. 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?

45. anonymous

take the derivative

46. anonymous

its not working for me

47. amistre64

check my work, and se if its right :)

48. anonymous

k the first term def works

49. anonymous

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

50. anonymous

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

51. 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

52. anonymous

im doing chain

53. anonymous

let me try again

54. anonymous

1 sec

55. amistre64

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

56. anonymous

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

57. anonymous

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

58. anonymous

thats the chain

59. amistre64

there you go :)

60. anonymous

i dont see how that makes the original

61. amistre64

cross your 4s, and whats left?

62. amistre64

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

63. anonymous

yea but its x^3 on the bottom

64. amistre64

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

65. anonymous

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

66. amistre64

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

67. anonymous

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

68. amistre64

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

69. anonymous

lol sorry

70. amistre64

lol .... well it does make it easier

71. anonymous

i wrote it wrong its my fault

72. amistre64

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

73. amistre64

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

74. anonymous

yes

75. amistre64

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

76. anonymous

k

77. amistre64

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

78. anonymous

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

79. amistre64

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

80. amistre64

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

81. anonymous

which form are you using

82. anonymous

thats right

83. anonymous

1/3/3=-8/3

84. anonymous

1/3-3 i mean

85. 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?

86. anonymous

let me take the derivative

87. amistre64

second term should be (-) not (+)

88. anonymous

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

89. 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

90. anonymous

oh ok

91. 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

92. anonymous

sounds good. lol it was so confusing im sorry

93. amistre64

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

94. anonymous

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

95. anonymous

?

96. 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...

97. anonymous

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

98. amistre64

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

99. anonymous

lol my brain isnt working forwards now

100. 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....

101. anonymous

oh ok

102. amistre64

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

103. anonymous

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

104. anonymous

never mind, you divide everything by x^3

105. anonymous

correct?

106. 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)

107. amistre64

yes, in essense, yes :)

108. anonymous

how does the 1 become x^3

109. anonymous

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

110. anonymous

omg i am acting dumb, sorry

111. amistre64

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

112. anonymous

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

113. amistre64

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

114. anonymous

yea it works out, complicated but works out

115. amistre64

its easier with a fresh pair of eyes :)

116. anonymous

yea thank you sooooooo much

117. amistre64

youre welcome :)

118. anonymous

Have a good night, ill ttyl