How Can I Solve These Calculus 2 Homework Problems?

Erbil
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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.
 
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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.
 
Thanks for reply.I'm waiting for more replies.Is there anybody who can help me?
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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