## 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?

1. freckles

Did you find the derivative to check yourself?

2. ganeshie8

To keep things simple, you may try u-substitution $$u = 6-x$$

3. anonymous

no i didn't find the derivative to check my answer, i would find the derivative of my problem right?

4. freckles

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

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

he might heard of it I don't know

7. anonymous

Im not sure i have, but i took the derivative and got x^2-12x+36x :)

8. freckles

x^3-12x^2+36x?

9. anonymous

yeah its just that i had x(x^2-12x+36)

10. freckles

11. anonymous

got it right! thanks!!

12. ganeshie8

how do you know  x(x^2-12x+36)  is same as the original function ?

13. ganeshie8

that might be a dumb q hmm

14. freckles

x(6-x)^2 think he expanded the square thingy first

15. anonymous

she* and yeah i did

16. ganeshie8

Ahh looks great she :D

17. freckles

oops sorry

18. freckles

do you want to see the sub thingy mentioned by ganeshie8 for fun?

19. anonymous

Its okay! sure is it easier than finding the derivative?

20. freckles

$\int\limits x (6-x)^2 dx \\ \text{ Let } u=6-x \\ \frac{du}{dx}=0-1 \\ \frac{du}{dx}=-1 \\ du=-1 dx \\ \text{ multiply both sides by -1 } \\ -du=dx \\ \text{ now recall if } u=6-x \text{ then } x=6-u \\ \text{ so we have } \\ \int\limits x (6-x)^2 dx=\int\limits (6-u)u^2 (-du) \\ =\int\limits (u-6)u^2 du$ so now you have less multiplication to do

21. freckles

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

22. anonymous

$x(6 − x)^2 = x(x^2-12x+36) = x^3-12x^2+36x$Then use power rule. No tricks needed. Your anti-derivative looks correct.

23. anonymous

Oh okay, I didnt understand the substitution part at first, but i get it now! thanks for explaining!

24. freckles

oops I didn't multiply correclty

25. freckles

my answer is off because when I did -6(u^2) I put -6u and so I integrated the wrong thingy

26. freckles

but that is sorta how works above if you can multiply correctly :p

27. ganeshie8

you should ask why on earth substitution is any better than your original method @gaba

28. ganeshie8

try this if you're loving antiderivatives : $\int x(6-x)^{9999999}\,dx = ?$

29. anonymous

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

im not sure i did it right though

31. ganeshie8

differentiate your answer and see if you get back the original function

32. ganeshie8

also what happened to the constant, C

33. anonymous

nvm I just did it again and got 4x^3/2/3 +6 -sinx +C

34. ganeshie8

you're doing calculus, that means you're not in highschool anymore time to use proper notation

35. ganeshie8

4x^ 3/2/3 +6 -sinx +C what does that even mean

36. anonymous

[(4x^3/2)/3] +6-sinx+C Btw you can take calculus in high school, just saying

37. ganeshie8

Much better, but still it is not mathematically correct

38. ganeshie8

pretty sure you meant [(4x^3/2)/3] +6(-sinx)+C

39. ganeshie8

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

why is it +6(sin(x)) and not -sin(x)?

41. freckles

$f(x)=2 \sqrt{x}+6 \cos(x) \\ \text{ or we can write } \\ f(x)=2 x^\frac{1}{2}+6 \cos(x)$ is this what you are playing with?

42. anonymous

yes!

43. freckles

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

forgot to put + some constant at the end there

45. anonymous

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

oh so you got derivative and antiderivative mixed up

47. freckles

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

but yeah antiderivative of 6cos(x) would be 6sin(x) since the derivative of 6sin(x) is 6cos(x)

49. freckles

the antiderivative of 6cos(x) would be 6sin(x)+C since the derivative of 6sin(x)+C is 6cos(x) *

50. freckles

the most general antiderivative *

51. anonymous

Okay I understand now! thanks a lot!!

52. freckles

np