- #1
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.
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.