Calculus 2 Based Homework

  • Thread starter Erbil
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  • #1
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1)
1.
y=x, y=sinx, x=-∏/4, x=∏/2 find the area between these limited curves.

2.
∫(upper function-(lower function)

3.
∫x-sinxdx(between x=-∏/4, x=0) + ∫sinx-xdx(between x=0,x=∏/2). Am I right? If it's okay,I don't have a problem in this integral.

2)
1.
∫sin(2x)/sin^2x(^=power)dx

2.
I'm not sure that this formula is for this equation.I just tried for solution.
sin^2x= 1/2(1-cos(2x))

3.∫sin(2x)/1/2(1-cos(2x))dx=2∫sin(2x)dx/(1-cos(2x)dx=2∫du/2/(1-cos(2x)=
=∫du/(1-cos(2x))

3)
1.
r(t)=t i + t^2/2 j + t^3/3 k, 0≤t≤6 find T,N,B for this function.

2.T(t) = r'(t)/|r'(t)| N(t) = T'(t)/|T'(t)| B=TXN

3.T(t) = i+2tj+t^2k/ I don't calculate this yet.I want to know what it is for ? Where we are use this vectors etc.In short,can somebody show it to me graphically?

4)
1.
Ʃ(it's starting from 0.And going to infinity.) n!(2x+3)^n
another information : it's a power series. What is a radial convergence R and interval convergence ?

2.lim n→∞ {an+1/an} {absolut value}

3.I have calculated this.And I found R=0, a=-3/2. It's just a point.
 

Answers and Replies

  • #2
938
9
1: I think you have upper and lower function mixed up
2: The function looks a bit like you might want to massage it to a form ∫f'(x)/f(x) dx. Perhaps try using trigonometry to the term upstairs.
3: It's a parametric curve in 3 dimensions. It's not very easy to graph, but if you write r=xi+yj+zk, then using the parametrization you are given, you can find what are y(x) and z(x), which you can use to find the shape of the curve.
 
  • #3
57
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Thanks for reply.I'm waiting for more replies.Is there anybody who can help me?
 
  • #4
57
0
1) I can't calculate it yet.
2)Yes you're right.So I found it. ln(sin^2x)+c
3)I don't try it yet.
4)I calculated it.

My homework is for tomorrow.Is there anybody,who can help me for area and vector question.
 

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