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anonymous

  • one year ago

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.) f(x) = x(6 − x)2 for my answer I got X^4/4-4x^3+18x^2+C is this right?

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  1. freckles
    • one year ago
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    Did you find the derivative to check yourself?

  2. ganeshie8
    • one year ago
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    To keep things simple, you may try u-substitution \(u = 6-x\)

  3. anonymous
    • one year ago
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    no i didn't find the derivative to check my answer, i would find the derivative of my problem right?

  4. freckles
    • one year ago
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    yeah you can differentiate your answer and see if it is x^3-12x^2+36x which is the form I think you put it in to integrate (by the way if you have heard of substitutions to make integrals like this nicer looking (or I mean easier) at ganeshie8's note )

  5. ganeshie8
    • one year ago
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    nvm if you haven't heard of u-substition before your work looks good, i presume you have used the formula for antiderivative of \(x^n\) to double check, differentiate the answer as freckles said

  6. freckles
    • one year ago
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    he might heard of it I don't know

  7. anonymous
    • one year ago
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    Im not sure i have, but i took the derivative and got x^2-12x+36x :)

  8. freckles
    • one year ago
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    x^3-12x^2+36x?

  9. anonymous
    • one year ago
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    yeah its just that i had x(x^2-12x+36)

  10. freckles
    • one year ago
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    and yes I already checked your answer integrating thingy but your problem said to also check your solution by differentiating so now you done both things your problem asked for

  11. anonymous
    • one year ago
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    got it right! thanks!!

  12. ganeshie8
    • one year ago
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    how do you know ` x(x^2-12x+36) ` is same as the original function ?

  13. ganeshie8
    • one year ago
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    that might be a dumb q hmm

  14. freckles
    • one year ago
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    x(6-x)^2 think he expanded the square thingy first

  15. anonymous
    • one year ago
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    she* and yeah i did

  16. ganeshie8
    • one year ago
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    Ahh looks great she :D

  17. freckles
    • one year ago
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    oops sorry

  18. freckles
    • one year ago
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    do you want to see the sub thingy mentioned by ganeshie8 for fun?

  19. anonymous
    • one year ago
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    Its okay! sure is it easier than finding the derivative?

  20. freckles
    • one year ago
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    \[\int\limits x (6-x)^2 dx \\ \text{ Let } u=6-x \\ \frac{du}{dx}=0-1 \\ \frac{du}{dx}=-1 \\ du=-1 dx \\ \text{ multiply both sides by -1 } \\ -du=dx \\ \text{ now recall if } u=6-x \text{ then } x=6-u \\ \text{ so we have } \\ \int\limits x (6-x)^2 dx=\int\limits (6-u)u^2 (-du) \\ =\int\limits (u-6)u^2 du\] so now you have less multiplication to do

  21. freckles
    • one year ago
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    \[=\int\limits (u-6)u^2 du =\int\limits (u^3-6u) du=\frac{u^4}{4}-3u^2+C \\ \text{ now remember } u=6-x \\ \text{ so you have } \\ =\frac{(6-x)^4}{4}-3(6-x)^2+C\] and it is less multiplication if you get to leave your answer like this :)

  22. anonymous
    • one year ago
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    \[ x(6 − x)^2 = x(x^2-12x+36) = x^3-12x^2+36x \]Then use power rule. No tricks needed. Your anti-derivative looks correct.

  23. anonymous
    • one year ago
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    Oh okay, I didnt understand the substitution part at first, but i get it now! thanks for explaining!

  24. freckles
    • one year ago
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    oops I didn't multiply correclty

  25. freckles
    • one year ago
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    my answer is off because when I did -6(u^2) I put -6u and so I integrated the wrong thingy

  26. freckles
    • one year ago
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    but that is sorta how works above if you can multiply correctly :p

  27. ganeshie8
    • one year ago
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    you should ask why on earth substitution is any better than your original method @gaba

  28. ganeshie8
    • one year ago
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    try this if you're loving antiderivatives : \[\int x(6-x)^{9999999}\,dx = ?\]

  29. anonymous
    • one year ago
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    I'm deff not loving them lol so Im working on this right now, Find the most general antiderivative of the function 2rootx+6cosx I got 4x^1/2+6-sinx

  30. anonymous
    • one year ago
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    im not sure i did it right though

  31. ganeshie8
    • one year ago
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    differentiate your answer and see if you get back the original function

  32. ganeshie8
    • one year ago
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    also what happened to the constant, C

  33. anonymous
    • one year ago
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    nvm I just did it again and got 4x^3/2/3 +6 -sinx +C

  34. ganeshie8
    • one year ago
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    you're doing calculus, that means you're not in highschool anymore time to use proper notation

  35. ganeshie8
    • one year ago
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    4x^ `3/2/3` +6 -sinx +C what does that even mean

  36. anonymous
    • one year ago
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    [(4x^3/2)/3] +6-sinx+C Btw you can take calculus in high school, just saying

  37. ganeshie8
    • one year ago
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    Much better, but still it is not mathematically correct

  38. ganeshie8
    • one year ago
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    pretty sure you meant [(4x^3/2)/3] +6(-sinx)+C

  39. ganeshie8
    • one year ago
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    differentiate that and see if you get back the original function (there is a mistake, so you wont get back the original function)

  40. anonymous
    • one year ago
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    why is it +6(sin(x)) and not -sin(x)?

  41. freckles
    • one year ago
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    \[f(x)=2 \sqrt{x}+6 \cos(x) \\ \text{ or we can write } \\ f(x)=2 x^\frac{1}{2}+6 \cos(x)\] is this what you are playing with?

  42. anonymous
    • one year ago
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    yes!

  43. freckles
    • one year ago
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    anyways ( ? )'=cos(x) the ?=sin(x) right? (sin(x))'=cos(x) so the antiderivative of cos(x) is sin(x)+C if I had ( ? )'=sin(x) the ?=-cos(x) (-cos(x))'=sin(x) so the antiderivative of sin(x) is -cos(x)+C now if you had ( ? )'=-cos(x) then ?=-sin(x) (-sin(x))'=-cos(x) so the antiderivative of -cos(x) is -sin(x)+C anyways you had find the antiderivative of 2x^(1/2) which looks like you were going for: 4x^(3/2)/3 +k but to find the antiderivative of 6cos(x) you need to recall what can you take derivative of that will give you 6cos(x) you can bring down the 6( and put the antiderivative of cos(x)) here )

  44. freckles
    • one year ago
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    forgot to put + some constant at the end there

  45. anonymous
    • one year ago
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    oh okay, is that to make sure I had it right I checked on mathway for the derivative of cos(x) and it said it was -sin(x) (which is what I thought it was before checking so it made sense) but i typed in +sin(x) and got it right, so thanks for explaining everything!

  46. freckles
    • one year ago
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    oh so you got derivative and antiderivative mixed up

  47. freckles
    • one year ago
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    but just so you know if it was taking derivative of 6cos(x) you would put -6sin(x) and not 6-sin(x)

  48. freckles
    • one year ago
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    but yeah antiderivative of 6cos(x) would be 6sin(x) since the derivative of 6sin(x) is 6cos(x)

  49. freckles
    • one year ago
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    the antiderivative of 6cos(x) would be 6sin(x)+C since the derivative of 6sin(x)+C is 6cos(x) *

  50. freckles
    • one year ago
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    the most general antiderivative *

  51. anonymous
    • one year ago
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    Okay I understand now! thanks a lot!!

  52. freckles
    • one year ago
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    np

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