## anonymous one year ago Assume that \sin(x) equals its Maclaurin series for all x. Use the Maclaurin series for sin(6 x^2) to evaluate the integral (integral [0, 0.78]) sin(6 x^2)dx . Your answer will be an infinite series. Use the first two terms to estimate its value.

1. anonymous

$\int\limits_{0}^{0.78}\sin(6x^2)dx$

2. anonymous

@mathstudent55

3. IrishBoy123

writing the Maclaurin sin expansion in terms of t: $$\large sin(t) = \Sigma_{0}^{\infty} \frac{(-1)^n }{(2n+1)! }\ t^{2n+1}$$ it seems they are happy for you to stuff $$t = 6 x^2$$ in and go from there to use an actual Mac series for $$sin(6 x^2)$$, you'd need to go $$f(0) = 0$$ $$f'(x) = (sin(6 x^2))' = ...$$ and $$f'(0) = ....$$ so $$x \ f'(0) = ...$$

4. anonymous

Just plug it in directly?

5. anonymous

What do you think @sourwing

6. IrishBoy123

"Just plug it in directly?" yes, that is what the question mandates for #1, plug it in term by terms and come up with a nice summation with a $$\Sigma$$ in it! for #2, generate a few terms and hope you get the first two quickly. [if you have a crack at the proper Mac expansion for that function, you will be busy for a while :p]

7. anonymous

So basically $\sum_{n = 0}^{\infty} \frac{ -1^n }{ (2n+1)! } * (6x^2)^{2n+1}$, and solve for n = 0 and n=1?

8. IrishBoy123

sounds good

9. anonymous

Well what do I do to estimate the value of the series?

10. anonymous

I ask because I still have an x term in my result

11. IrishBoy123

if you have an 'x' term in your result, well that is a good thing. :p

12. anonymous

Okay lol. Sorry I'm clueless on the procedure here. Should I be evaluating the limit at both of the solutions? They are N_0 = -6x^2 and N_1 = -216x^2

13. IrishBoy123

now, forget about the $$\Sigma$$ and just evaluate the first 2 terms you are not summing, you are just using the first 2 terms of the series make sense?!?! you need x terms because you have to integrate

14. IrishBoy123

nothing to do with limits, either. just knock out the first 2 terms of the [bogus] expansion

15. anonymous

So get the first two terms in the series, but how do I use that to estimate the value of the series?

16. IrishBoy123

|dw:1439852188969:dw|

17. IrishBoy123

|dw:1439852553035:dw|

18. anonymous

$6x^2-x^5$??

19. IrishBoy123

|dw:1439852806038:dw|

20. anonymous

-1/6 * 6x^5

21. anonymous

-6x^5/6

22. anonymous

-x^5

23. IrishBoy123

you are *not* summing a series, or *valuing* a series you are hacking an integral; by approximating the integrand to a hack of its Mac expansion

24. anonymous

brb I have to go feed goats

25. anonymous

@Kainui @Hero I'm sorry, but I still don't understand how to "hack the integral"? I'm still not quite getting what IrishBoy is trying to have me do here. Could you or one of the other moderators help here?

26. anonymous

Maclaurin series of $$sin(6x^2)$$ is $$sin(6x^2) =\sum_{n=0}^{\infty}\dfrac{(-1)^n (6x^2)^{2n+1}}{(2n+1)!}$$ $\int_0^{0.78} sin(6x^2)dx=\int_0^{0.78}\sum_{n=0}^{\infty}\dfrac{(-1)^n(6x^2)^{2n+1}}{(2n+1)!}dx\\=\sum_{n=0}^{\infty}\int_0^{0.78}\dfrac{(-1)^n(6x^2)^{2n+1}}{(2n+1)!} dx$ This integral is easy to take, right? After take out the integral, you just calculate the first 3 term of it (approximate theorem) and you get the answer.

27. anonymous

oh, your problem asks for the first 2 terms, not 3 as I suggested. That is easier.!!

28. anonymous

I'm sorry but nothing about this seems easy to me. I hate to keep bothering you but is there any way you can force me to understand this? @Crazyandbeautiful

29. anonymous

OH, I assumed you know the formula of Maclaurin series of sin (6x^2) If you don't know it, you can derive it from Maclaurin series of sinx, ok?

30. anonymous

$$sin x =\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}$$ ok?

31. anonymous

Okay

32. anonymous

so, just replace x by 6x^2 to get Maclaurin series of sin(6x^2)

33. anonymous

ok?

34. anonymous

Alright

35. anonymous

now, you want to take int of sin (6x^2) from 0 to 0.78 , right? Just take integral both sides.

36. anonymous

The left hand side is $$\int_0^{0.78} sin(6x^2)dx$$

37. anonymous

The right hand side is integral of the sum (...) right?

38. anonymous

Yes or NO? where don't you get?

39. anonymous

I'm just writing this in my notes as we go, sorry

40. anonymous

Yes

41. anonymous

And we know that integral of sum = sum of integral, right?

42. anonymous

Okay

43. anonymous

hence the right hand side is $\sum_{n=0}^{\infty} \int_0^{0.78}(-1)^n\dfrac{(6x^2)^{2n+1}}{(2n+1)!}$ ok?

44. anonymous

Okay, I'm with you

45. anonymous

Now, calculate the integral. Can you?

46. anonymous

I can try.

47. anonymous

But what do I do with the fact that this has both n and x in it?

48. anonymous

Don't forget $$(6x^2)^{2n+1}= 6^{2n+1}*x^{4n+2}$$

49. anonymous

hey, just x is the variable, n is a number!! nothing to do with n, take it as constant

50. anonymous

you take integral w.r.t.x (dx) right? how n works as variable?

51. anonymous

We will use n for the far left summation, not for integral.

52. anonymous

Okay

53. anonymous

It turns to $\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!}\int_0^{0.78} x^{4n+2}dx$

54. anonymous

ok?? take integral, what do you get?

55. anonymous

|dw:1439861021631:dw|

56. anonymous

$\frac{ x ^{4n+3} }{ 4n+3 }$???

57. anonymous

58. anonymous

Sorry, evaluated from 0 to 0.78

59. anonymous

ok, =?

60. anonymous

Just sub .78 and 0 for x and subtract?

61. anonymous

0 for x =0, just plug x =0.8 in, ok?

62. anonymous

oh, 0.78

63. anonymous

So it's just $\frac{ .78^{4n+3} }{ 4n+3 }$ ??

64. anonymous

So, now you have $\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}$

65. anonymous

ok ?

66. anonymous

Yes

67. anonymous

You need just 2 terms!! easy!! n =0, plug in, n =1, plug in add them together, done.

68. anonymous

That's it? Just add the resulting two terms together?

69. anonymous

Yes, dat sit. If you want, you can add you and me to get 4 terms. :)

70. anonymous

71. anonymous

Working on it now

72. anonymous

-0.745188?

73. anonymous

I got 0.0457318875

74. anonymous

ok, let's do together

75. anonymous

I'm absolutely sure that you're right, I just did mine mostly in my head...

76. anonymous

$\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}$ $(-1)^0\dfrac{6^{2*0+1}}{(2*0+1)!} *\dfrac{0.78^{4*0+3}}{4*0+3}+(-1)^1\dfrac{6^{2*1+1}}{(2*1+1)!} *\dfrac{0.78^{4*1+3}}{4*1+3}$

77. anonymous

$-1^{0}*\frac{ 6^{2(0)+1} }{ (2(0)+1)! }*\frac{ 0.78^{4(0)+3} }{ 4(0)+3 }$

78. anonymous

Oh you're much faster at this than I am

79. anonymous

hahaha... I just copy and paste the previous post, then change the numbers. hehehe

80. anonymous

Well for n = 0, term by term, the first one would equal 1 * 6/1* .78^3/3 which I said totals 0.949104

81. anonymous

and your first term = $$\dfrac{6}{1!}\dfrac{0.78^3}{3}$$ right? = $$2*0.78^3=0.949104$$ Yes,

82. anonymous

The second term is ??

83. anonymous

For n = 1 I said -1* 6^3/6 * .78^7/7 which I thought should be -0.90337211

84. anonymous

$$-\dfrac{6^3}{3!}\dfrac{0.78^7}{7}=-0.9033721125$$ Yes, again,

85. anonymous

add them together , you get??

86. anonymous

0.045732

87. anonymous

yup

88. anonymous

So I did just mess something up in my head the first time.

89. anonymous

easy, right?

90. anonymous

Thank you so much Crazyandbeautiful, you are literally the most beautiful person in the world to me right now

91. anonymous

haahaha... I'm not beautiful, I am handsome. hahaha.... jk

92. anonymous

lol you're probably going to be the deciding factor in how well I do in my test next week actually, because I clearly had no idea what I was doing.

93. anonymous

Handsome, beautiful, the actual envisionment of God, whatever you are dude you saved my life here

94. anonymous

You got the method, right? don't scare me. I spent more than 1h to explain and you said you don't know what you are doing. it is a real scaring.

95. anonymous

I had no idea BEFORE this chat string

96. anonymous

Believe me, I was taking notes

97. anonymous

ok, good luck

98. anonymous

Thanks again