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 Calculus study realm

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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 16:40

e) ∫(0 to 6) |x - 2| dx = ∫(0 to 2) -(x - 2) dx + ∫(2 to 6) (x - 2) dx = (-x^2 /2 + 2x)|(0 to 2) + (x^2 /2 - 2x)|(2 to 6) = -2 + 4 +18 - 12 - 2 + 4 = 10
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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 16:42

-_-lll Don't forget that I am the one who alway let you copy the answers for the assignments.

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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 16:43

Oh, evil-mashimaro, can you post your answers for the assignments? 6

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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 16:44

1 That's gonna take me forever! No way! Ask Spirit. He has all my answers.

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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 17:18

I don't know where did I put my assignments. And I have all the exercises sheets to cover. Go ahead, do the assignments.

4) Let the function y=f(x) be defined by f(x) = {x^2 - 1, if -1≤x≤2 | 4 - x, if 2<x≤4}. If y=g(x) is defined by g(x) = ∫(-1 to x) f(t) dt, using Riemann sums with Δx=1 and RHS rectangles, find an approximate graph of y=g(x) on the interval -1≤x≤4.

g(-1) = ∫(-1 to -1) f(t) dt = 0

g(0) = ∫(-1 to 0) f(t) dt = f(0) (1) = -1

g(1) = ∫(-1 to 0) f(t) dt + ∫(0 to 1) f(t) dt = -1 + f(1) (1) = -1

g(2) = ∫(-1 to 1) f(t) dt + ∫(1 to 2) f(t) dt = -1 + f(2) (1) = 2

g(3) = ∫(-1 to 2) f(t) dt + ∫(2 to 3) f(t) dt = 2 + f(3) (1) = 3

g(4) = ∫(-1 to 3) f(t) dt + ∫(3 to 4) f(t) dt = 3 + f(4) (1) = 3

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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 17:26

5) Let y=g(x) be defined by g(x) = ∫(0 to x) sint^3 dt, find dg(x)/dx.

dg(x)/dx = sinx^3

6) Let y=g(x) be defined by g(x) = ∫(x to 1) (t^4 + 4t)^(1/3) dt, find dg(x)/dx.

dg(x)/dx = -(x^4 + 4x)^(1/3)

7) Let y=g(x) be defined by g(x) = ∫(3 to x^3) t^2 /(t^3 + 1) dt, find dg(x)/dx.

dg(x)/dx = (x^3)^2 /((x^3)^3 + 1) (3x^2) = 3x^8 /(x^9 +1)
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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 23:05

Transformation

1) ∫(1 - √x)^2 dx = ∫(1 - 2√x + x) dx = x - 4/3 x^(3/2) + x^2 /2 + C
2) ∫(1 - √x)x^(1/3) dx = ∫(x^(1/3) + x^(5/6)) dx = 3/4 x^(4/3) + 6/11 x^(11/6) + C
3) ∫(1 + x)^2 /√x dx = ∫(1/√x + 2√x + x^(3/2)) dx = 2√x + 4/3x^(3/2) + 2/5 x^(5/2) + C
4) ∫(tanx + secx)^2 dx = ∫((tanx)^2 + 2 tanx secx + (secx)^2) dx = ∫((secx)^2 - 1 + 2 tanx secx + (secx)^2) dx = tanx - x + 2secx + tanx + C = 2tanx + 2secx - x + C
5) ∫(1 + tanx)^2 dx = ∫(1 + 2tanx + (tanx)^2) dx = ∫(1 + 2tanx + (secx)^2 - 1) dx = 2ln|secx| + tanx + C
6) ∫(sinx + cosx)/cosx dx = ∫(tanx + 1) dx = x + ln|secx| + C
7) ∫dx/(1 + cosx) = ∫(1 - cosx)/(1 - (cosx)^2) dx = ∫((cscx)^2 - cscx cotx) dx = cscx - cotx + C
8) ∫dx/(1 - sinx) = ∫(1 + sinx)/(1 - (sinx)^2) dx = ∫((secx)^2 + tanx secx) dx = tanx + secx + C
9) ∫dx/(secx + 1) = ∫(secx - 1)/((secx)^2 - 1) dx = ∫(cotx cscx - (cotx)^2) dx = ∫(cotx cscx - ((cotx)^2 + 1 - 1) dx = ∫(cotx cscx - (cscx)^2 + 1) dx = cotx - cscx + x + C
10) ∫(e^x + 1)/e^x dx = ∫(1 + e^-x) dx = x - e^x + C
11) ∫(1 + x)/(1 + x^2) dx = ∫(1/(1 + x^2) + x/(1 + x^2)) dx = arctanx + ln|1 + x^2|/2 + C
12) ∫dx/(1 + e^x) = ∫(1 + e^x - e^x)/(1 + e^x) dx = ∫(1 - e^x /(1 + e^x)) dx = x - ln|1 + e^x| + C
13) ∫(x + 2)/(x + 1) dx = ∫(x + 1 + 1)/(x + 1) dx = ∫(1 + 1/(x + 1)) dx = x + ln|x + 1| + C
14) ∫(x^2 + 1)/(x + 3) dx = (after long division) ∫(x - 3 + 10/(x + 3)) dx = x^2 /2 - 3x + 10ln|x + 3| + C
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MessageSujet: Re: Calculus study realm    Dim 15 Mai 2011 - 23:54

Substitution

1) ∫dx/(2x - 3) = ∫du/2u = ln|u|/2 = ln|2x - 3|/2 + C
Let u = 2x - 3
du = 2

2) ∫x√(x - 4) dx = ∫(u + 4)√u du = ∫(u^(3/2) + 4√u) du = 2/5 u^(5/2) + 8/3 u^(3/2) + C = 2/5 (x - 4)^(5/2) + 8/3 (x - 4)^(3/2) + C
Let u = x - 4
du = 1

3) ∫(e^x + 1)^3 e^x dx = ∫u^3 du = u^4 /4 + C = (e^x + 1)^4 /4 + C
Let u = e^x + 1
du = e^x

4) ∫(1 + x^2)^(1/3) x dx = ∫u^(1/3) /2 du = 3/8 u^(4/3) + C = 3/8 (1 + x^2)^(4/3) + C
Let u = 1 + x^2
du = 2x

5) ∫x^2 /(x^3 + 2)^(1/4) dx = ∫du /3u^(1/4) = 4/9 u^(3/4) + C = 4/9 (x^3 + 2)^(3/4) + C
Let u = x^3 + 2
du = 3x^2

6) ∫8x^2 /(x^3 + 4)^3 dx = ∫8/3u^3 dx = -4/3u^2 + C = -4/3(x^3 + 4)^2 + C
Let u = x^3 + 4
du = 3x^2

7) ∫sin(3x)/(cos(3x))^2 dx = ∫du/3u^2 = -1/3u + C = -1/3cos(3x) + C = -sec(3x)/3 + C
Let u = cos(3x)
du = -3sin(3x)

8) ∫(sin(4x))^2 cos(4x) dx =∫u^2 /4 du = u^3 /12 + C = (sin(4x))^3 /12 + C
Let u = sin(4x)
du = 4cos(4x)

9) ∫sin(2x) ((sinx)^2 + 1)^4 dx = ∫sin(2x) ((1/2 - 1/2 cos(2x)) + 1)^4 dx = ∫sin(2x) (3/2 - 1/2 cos(2x))^4 dx = ∫u^4 du = u^5 /5 + C = (3/2 - 1/2 cos(2x))^5 /5 + C
Let u = 3/2 - 1/2 cos(2x)

10) ∫e^(3cos(2x)) sin(2x) dx = ∫-e^u /6 du = -e^u /6 + C = -e^(3cos(2x)) /6 + C
Let u = 3cos(2x)
du = -6sin(2x)
du = sin(2x)

11) ∫x cotx^2 dx = ∫cotu /2 du = ln|sinu|/2 + C = ln|sinx^2|/2 + C
Let u = x^2
du = 2x

12) ∫(sec√x)/√x dx = ∫2secu du = 2ln|secu + tanu| + C = 2ln|sec√x + tan√x| + C
Let u = √x
du = 1/2√x

13) ∫x^5 √(x^3 + 3) dx = ∫(u - 3)/3 √u du = ∫(u^(3/2) /3 - √u) du = 2/15 u^(5/2) - 2/3 u^(3/2) + C = 2/15 (x^3 + 3)^(5/2) - 2/3 (x^3 + 3)^(3/2) + C
Let u = x^3 + 3
du = 3u^2

14) ∫(secx/(1 - tanx)^2 dx = ∫-du/u^2 = 1/u + C = 1/(1 - tanx) + C
Let u = 1 - tanx
du = -(secx)^2

15) ∫x/(a + bx^2) dx = ∫du/2bu = ln|u|/2b + C = ln|a + bx^2|/2b + C
Let u = a + bx^2
du = 2bx

16) Show that ∫(0 to 3) f(3 - x) dx = ∫(0 to 3) f(x) dx

Let u = 3 - x
du = -1

∫(0 to 3) f(3 - x) dx = ∫(3 to 0) -f(u) du = ∫(0 to 3) f(u) du = ∫(0 to 3) f(x) dx
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 0:01

I can't take it anymore! It's too painful! O, pity! 039
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 14:28

Come on, oppa. I won't pass the test without you. 010

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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 14:30

In that case, ok. 016
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 14:34

XD Why does it sounds so weird?

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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 19:02

Trig Sub

1) ∫dx/√(2 - x^2) = ∫√2cosθ/√(2 - 2(sinθ)^2) dθ = ∫dθ = θ = arcsin(x/√2) + C
Let x =√2sinθ
dx = √2cosθ dθ

2) ∫x^2 /(x^2 + 4) dx = ∫2(secθ)^2 4(tanθ)^2 /(4(tanθ)^2 + 4) dθ = ∫2(tanθ)^2 dθ = 2∫((secθ)^2 -1) dθ = 2tanθ - 2θ + C = x - 2arctan(x/2) + C
Let x = 2tanθ
dx = 2(secx)^2 dθ

3) ∫dx/(x^2 √(x^2 - 4)) = ∫2secθtanθ/((2secθ)^2 √(4(secθ)^2 - 4)) dθ = ∫ dθ/4secθ = 1/4 ∫cosθ dθ = 1/4 sinθ = √(x^2 - 4) /4x + C
Let x = 2secθ; secθ = x/2; sinθ = √(x^2 - 4) /x
dx = 2secθtanθ dθ

4) ∫dx/(x√(4x^2 - 9)) = ∫3/2 secθtanθ/(3/2 secθ√(9(secθ)^2 - 9)) dθ = ∫dθ/3 = θ/3 + C = 1/3 arcsec(2x/3) + C
Let 2x = 3secθ
dx = 3/2 secθtanθ dθ

5) ∫√(9 - 4x^2)/x dx = ∫3/2 cosθ √(9 - 9(sinθ)^2)/(3/2sinθ) dθ = ∫3(cosθ)^2 /sinθ dθ = ∫3(1-(sinθ)^2)/sinθ dθ = ∫3(cscθ - sinθ) dθ = 3ln|cscθ - cotθ| + 3cosθ + C = 3ln|(3 - √(9 - 4x^2))/2x| + √(9 - 4x^2) + C
Let 2x = 3sinθ; sinθ = 2x/3; cosθ = √(9 - 4x^2)/3
dx = 3/2 cosθ dθ

6) ∫secxtanx/(9 + 4(secx)^2) dx = ∫3/2 (secθ)^2 /(9 + 9(tanθ)^2) dθ = ∫3/2 (secθ)^2 /(9 + 9(tanθ)^2) dθ = 3/2 ∫1/9 dθ = 1/6 θ + C = 1/6 arctan(2/3 secx) + C
Let 2secx = 3tanθ
secxtanx dx = 3/2 (secθ)^2 dθ

7) ∫(3x - 7)/(x^2 - 6x + 10) dx = ∫(3x - 7)/((x - 3)^2 +1) dx = ∫(3(tanθ + 3) - 7)(sexθ)^2 /((tanθ)^2 +1) dθ = ∫(3tanθ + 2) dθ = 3ln|secθ| + 2θ + C = 3/2 ln|sec(x^2 - 6x + 10)| + 2arctan(x - 3) + C
Let x - 3 = tanθ
dx = (sexθ)^2 dθ

8) ∫(2x + 3)/(9x^2 - 12x + 8) dx = ∫(2x + 3)/((3x - 2)^2 + 4) dx = ∫2/3 (secθ)^2 (2(2/3 tanθ + 2/3) + 3)/(4(tanθ)^2 + 4) dθ = ∫2/3 (secθ)^2 (4/3 tanθ + 13/3)/4(secθ)^2 dθ = ∫(4/3 tanθ + 13/3)/6 dθ = ∫(2/9 tanθ + 13/18) dθ = 2/9 ln|secθ| + 13/18 θ + C = 2/9 ln|√(9x^2 - 12x + 8)/2| + 13/18 arctan(3/2 x - 1) + C = 1/9 ln|(9x^2 - 12x + 8)| + 13/18 arctan(3/2 x - 1) + C
Let (3x - 2) =2tanθ
dx = 2/3 (secθ)^2 dθ

9) ∫dx/(x^2 + 10x + 30) = ∫dx/((x + 5)^2 + 5) = ∫√5(secθ)^2/(5(tanθ)^2 + 5) dθ = ∫dθ/√5 = θ/√5 + C = arctan((x + 5)/√5) /√5 + C = √5/5 arctan(√5(x + 5)/5) + C
Let (x + 5) = √5tanθ
dx = √5(secθ)^2 dθ

10) ∫dx/√(28 - 12x - x^2) = ∫dx/√(64 - (6 + x)^2) = ∫8cosθ/√(64 - 64(sinθ)^2) dθ = ∫dθ = θ + C = arcsin(3/4 +x/8) + C
Let (6 + x) = 8sinθ
dx = 8cosθ dθ

11) ∫dx/x√(x^2 - 2x) = ∫dx/x√((x - 1)^2 - 1) =∫secθtanθ/(secθ + 1)√((secθ)^2 - 1)dθ = ∫secθ/(secθ + 1) (secθ - 1)/(secθ - 1) dθ = ∫((secθ)^2 - secθ)/(tanθ)^2 = ∫((cscθ)^2 - cscθcotθ) dθ = -cotθ + cscθ + C = -1/√(x^2 - 2x) + (x - 1)/√(x^2 - 2x) + C = (x - 2)/√(x^2 - 2x) + C
Let (x - 1) = secθ
dx = secθtanθ dθ

12) ∫dx/(36 + x^2)^2 = ∫6(secθ)^2 /(36 + 36(tanθ)^2)^2 dθ = ∫6(secθ)^2 /(36 + 36(tanθ)^2)^2 dθ = ∫(cosθ)^2 /216 dθ = ∫(1/2 + 1/2 cos2θ)/216 dθ = 1/216 (1/2 θ + 1/4 sin2θ) + C = 1/216 (1/2 arctan(x/6) + 1/4 sin2θ) + C = 1/432 arctan(x/6) + 1/432 sinθcosθ + C = 1/432 arctan(x/6) + 1/432 x/√(36 + x^2) 6/√(36 + x^2) + C = 1/432 arctan(x/6) + x/72(36 + x^2) + C
Let x = 6tanθ
dx = 6(secθ)^2 dθ

13) ∫dx/(x^4 √(x^2 - 3)) = ∫√3secθtanθ/(9(secθ)^4 √(3(secθ)^2 - 3)) dθ = ∫(cosθ)^3 /9 dθ = ∫cosθ(1 - (sinθ)^2) /9 dθ = sinθ /9 - (sinθ)^3 /27 + C = √(x^2 - 3)/9x - ((x^2 - 3)/x)^3 /27 + C = 3x^2√(x^2 - 3)/27x^3 - ((x^2 - 3)^3/27x^3) + C = (3x^2√(x^2 - 3) - √(x^2 - 3)^3)/27x^3 + C = (3x^2 - (x^2 - 3))√(x^2 - 3)/27x^3 + C = (2x^2 + 3)√(x^2 - 3)/27x^3 + C
Let x = √3secθ
dx = √3secθtanθ dθ

14) ∫(4 + x^2)^2 /x^3 dx = ∫(16 + 8x^2 + x^4)/x^3 dx = ∫(16/x^3 + 8/x + x) dx = -8/x^2 + 8ln|x| + x^2 /2 + C
No trig sub

15) ∫dx/√(x^2 + a^2) = ∫a(secθ)^2 /√(a^2 (tanθ)^2 + a^2) dθ = ∫secθ dθ = ln|secθ + tanθ| + C = ln|√(x^2 + a^2)/a + x/a| + C = ln|√(x^2 + a^2) + x| + C
Let x = a tanθ
dx = a(secθ)^2 dθ
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 19:05

Can you just do the 5 last questions for each integration exercise sheet?

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Dernière édition par JellyFish le Lun 16 Mai 2011 - 19:16, édité 1 fois
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 19:06

Why didn't you tell me before?! 07
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 20:31

Further Substitution

9) ∫dx/(x - x^(2/3)) = ∫dx/x^(2/3)(x^(1/3) - 1) = ∫dx/x^(2/3)(x^(1/3) - 1) = ∫3/u du = 3ln|u| + C = 3ln|x^(1/3) - 1| + C
Let u = x^(1/3) - 1
du = 1/3x^(2/3) du

10) ∫dx/x(1 - x^(1/4)) = -∫4du/u(1 - u) = -4∫du/u(1 - u) = -4∫(1/u + 1/(1 - u)) du = -4ln|u| + 4ln|1 - u| + C = 4ln|(1-u)/u| + C = 4ln|x^(1/4) /(1 - x^(1/4))| + C
Let u = 1 - x^(1/4)
du = -1/4x^(3/4) dx

Consider 1/u(1 - u) = A/u + B/(1 - u) = (A(1 - u) + Bu)/u(1 - u)
For u = 1, B = 1
For u = 0, A = 1

11) ∫(√(x + 1) - 1)/(√(x + 1) + 1) dx = ∫(√(x + 1) + 1 - 2)/(√(x + 1) + 1) dx = ∫(1 - 2/(√(x + 1) + 1)) dx = ∫2u(1 - 2/(u + 1)) du = ∫(2u - 4u/(u + 1)) du = ∫(2u - (4u + 4 - 4)/(u + 1)) du = ∫(2u - 4 + 4/(u + 1)) du = u^2 - 4u + 4ln|u + 1| + C = x - 4√(x + 1) + 4ln|√(x + 1) + 1| C
Let u^2 = x + 1
2u du = dx

12) ∫dx/√(1 + √x) = ∫2u/√(1 + u) du = ∫4v(v^2 - 1)/v dv = ∫(4v^2 - 4) dv = 4/3 v^3 - 4v + C = 4/3 (1 + u)^(3/2) - 4√(1 + u) + C = 4/3 (1 + √x)^(3/2) - 4√(1 + √x) + C
Let u^2 = x
2u du = dx

Let v^2 = 1 + u
2v dv = du

13) ∫dx/(c+fx)√(a + bx) = ∫2u/b(c+f(u^2 - a)/b)u du = ∫2/(bc + fu^2 - fa) du = ∫2/((bc - fa) + fu^2) = ∫2√(bc/f - a)(secθ)^2 /((bc - fa) + (bc - fa)(tanθ)^2) dθ = ∫2√(bc/f - a)/(bc - fa) dθ = 2/√(f(bc - fa)) θ + C = 2/√(f(bc - fa)) arctan(√fu/√(bc - fa)) + C = 2/√(f(bc - fa)) arctan(√(f(a + bx))/√(bc - fa)) + C

Let u^2 = a + bx
2u du = b dx

Let √fu = √(bc - fa)tanθ
du = √(bc/f - a)(secθ)^2 dθ
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 21:44

Power of trig
11) ∫√cotx (cscx)^4 dx = ∫√cotx ((cotx)^2 + 1) (cscx)^2 dx = ∫(cotx)^(5/2) (cscx)^2 + √cotx (cscx)^2 dx = -2/7 (cotx)^(7/2) - 2/3 (cotx)^(3/2) + C

12) ∫sin2x sin4x dx = ∫sin2x (2sin2x cos2x) dx = ∫2(sin2x)^2 cos2x dx = (sin2x)^3 / 3

13) ∫sin4x (cos3x)^2 dx = ∫sin4x (1/2 + 1/2 cos6x) dx = ∫(1/2 sin4x + 1/2 sin4x cos6x) dx = ∫(1/2 sin4x + 1/2 (1/2 (sin(-2x) + sin10x)) dx = ∫(1/2 sin4x - 1/4 sin2x + 1/4 sin10x) dx = -1/8 cos4x + 1/8 cos2x - 1/40 cos10x + C

14) ∫(tan(ax))^n dx = ∫(tan(ax))^2 (tan(ax))^(n-2) dx = ∫((sec(ax))^2 - 1)(tan(ax))^(n-2) dx = ∫(sec(ax))^2 (tan(ax))^(n-2) - (tan(ax))^(n-2) dx = (tan(ax))^(n-1) /a(n-1) - ∫(tan(ax))^(n-2) dx + C

15) ∫(sinx)^n dx = -cosx(sinx)^(n-1) + ∫(n-1)(sinx)^(n-2) (cosx)^2 dx = -cosx(sinx)^(n-1) + (n-1)∫(sinx)^(n-2) (1 - (sinx)^2) dx = -cosx(sinx)^(n-1) + (n-1)∫((sinx)^(n-2) - (sinx)^n) dx

n∫(sinx)^n dx = -cosx(sinx)^(n-1) + (n-1)∫(sinx)^(n-2)dx
∫(sinx)^n dx = -cosx(sinx)^(n-1) /n + (n-1)/n ∫(sinx)^(n-2)dx


Let f(x) = -cosx, f'(x) = sinx
g(x) = (sinx)^(n-1), g'(x) = (n-1)(sinx)^(n-2)cosx
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MessageSujet: Re: Calculus study realm    Lun 16 Mai 2011 - 22:59

Integration by part

11) ∫sin(lnx) dx = sin(lnx) x - ∫cos(lnx) dx + C = sin(lnx) x - cos(lnx) x - ∫sin(lnx) dx + C

f(x) = sin(lnx)
f'(x) = cos(lnx)/x

g(x) = x
g'(x) = 1

--

f(x) = cos(lnx)
f'(x) = -sin(lnx)/x

g(x) = x
g'(x) = 1

2∫sin(lnx) dx = sin(lnx) x - cos(lnx) x + C
∫sin(lnx) dx = sin(lnx) x/2 - cos(lnx) x/2 + C



12) ∫e^(2x) cos(3x) dx = cos(3x) e^(2x) /2 + 3/2 ∫sin(3x) e^(2x) dx + C = cos(3x) e^(2x) /2 + 3/2 (sin(3x) e^(2x) /2 - 3/2 ∫cos(3x) e^(2x) dx) + C

f(x) = cos(3x)
f'(x) = -3sin(3x)

g(x) = e^(2x) /2
g'(x) = e^(2x)

--

f(x) = sin(3x)
f'(x) = 3cos(3x)

g(x) = e^(2x) /2
g'(x) = e^(2x)

13/4 ∫e^(2x) cos(3x) dx = cos(3x) e^(2x) /2 + 3/2 sin(3x) e^(2x) /2 + C
∫e^(2x) cos(3x) dx = 2/13 cos(3x) e^(2x) + 3/13 sin(3x) e^(2x) + C




13) ∫e^(ax) sin(bx) dx = -e^(ax) cos(bx) /b + a/b ∫e^(ax) cos(bx) dx + C = -e^(ax) cos(bx) /b + a/b (e^(ax) sin(bx) /b - a/b ∫e^(ax) sin(bx) dx) + C

f(x) = e^(ax)
f'(x) = ae^(ax)

g(x) = -cos(bx) /b
g'(x) = sin(bx)

--

f(x) = e^(ax)
f'(x) = ae^(ax)

g(x) = sin(bx) /b
g'(x) = cos(bx)

(1 + (a/b)^2) ∫e^(ax) sin(bx) dx = -e^(ax) cos(bx) /b + a/b^2 e^(ax) sin(bx) + C
∫e^(ax) sin(bx) dx = 1/(1 + (a/b)^2) (-e^(ax) cos(bx) /b + a/b^2 e^(ax) sin(bx)) + C = b^2 /(b^2 + a^2) (-e^(ax) cos(bx) /b + a/b^2 e^(ax) sin(bx)) + C = -e^(ax) cos(bx) b /(b^2 + a^2) + e^(ax) sin(bx)) a /(b^2 + a^2) + C



14) ∫x^n lnx dx = lnx x^(n+1) /(n+1) - 1/(n+1) ∫x^n dx + C = lnx x^(n+1) /(n+1) - 1/(n+1) x^(n+1) /(n+1) + C = lnx x^(n+1) /(n+1) - x^(n+1) /(n+1)^2 + C

f(x) = lnx
f'(x) = 1/x

g(x) = x^(n+1) /(n+1)
g'(x) = x^n

15) ∫x^m (x^2 + a)^(1/n) dx = x^(m-1) (x^2 + a)^(1 + 1/n) /(2 + 2/n) - (m-1)/(2 + 2/n) ∫x^(m-2) (x^2 + a)^(1 + 1/n) dx + C = n/(2+2n) x^(m-1) (x^2 + a)^(1 + 1/n) - (m-1)n/(2 + 2n) ∫x^(m-2) (x^2 + a)^(1 + 1/n) dx + C

f(x) = x^(m-1)
f'(x) = (m-1)x^(m-2)

g(x) = (x^2 + a)^(1 + 1/n) /2(1 + 1/n) = (x^2 + a)^(1 + 1/n) /(2 + 2/n)
g'(x) = x(x^2 + a)^(1/n)
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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 0:38

Partial fraction

11) ∫(4x^2 + 13x - 9)/(x^3 + 2x^2 - 3x) dx = ∫(4x^2 + 13x - 9)/x(x+3)(x-1) dx

Consider (4x^2 + 13x - 9)/x(x+3)(x-1) = A/x + B/(x+3) + C/(x-1) = (A(x-1)(x+3) + Bx(x-1) + Cx(x+3))/x(x-1)(x+3) = (A(x^2 + 2x - 3) + B(x^2 - x) + C(x^2 + 3x))/x(x-1)(x+3)

x^2: A + B + C = 4; B = 4 - A - C
x: 2A - B + 3C = 13
1: -3A = -9; A = 3

13 = 2(3) - (4 - 3 - C) + 3C = 5 + 4C
C = 2
B = -1

∫(4x^2 + 13x - 4)/x(x+3)(x-1) dx = 3∫dx/x -∫dx/(x-3) + 2∫dx(x-1) = 3ln|x| - ln|x-3| + 2ln|x-1| + C = ln|(x^3 (x-1)^2)/(x-3)| + C

12) Consider (3x^3 -18x^2 + 29x - 4)/(x + 1)(x - 2)^3 = A/(x + 1) + B/(x - 2) + C/(x - 2)^2 + D/(x - 2)^3 = (A(x - 2)^3 + B(x + 1)(x - 2)^2 + C(x + 1)(x - 2) + D(x + 1))/(x + 1)(x - 2)^3 = (A(x^3 - 6x^2 + 12x - 8) + B(x^3 - 3x^2 + 4) + C(x^2 - x - 2) + D(x + 1))/(x + 1)(x - 2)^3

x^3: A + B = 3; A = 3 - B
x^: -6A - 3B + C = -18; C = 3B + 6(3 - B) - 18 = -3B
x: 12A - C + D = 29; D = 29 - 12(3 - B) + (-3B) = 9B - 7
1: -8A + 4B - 2C + D = -4

-8(3 - B) + 4B - 2(-3B) + (9B - 7) = -4
27B = 27
B = 1
A = 2
C = -3
D = 2

∫(3x^3 -18x^2 + 29x - 4)/(x + 1)(x - 2)^3 dx = 2∫dx/(x + 1) + ∫dx/(x - 2) - 3∫dx/(x - 2)^2 + 2∫dx/(x - 2)^3 = 2ln|x-2| + ln|x-2| + 3/(x-2) - 3/2(x-2) + C




13) Consider (5x^3 - 3x^2 + 7x - 3)/(x^2 + 1)^2 = (Ax + B)/(x^2 + 1) + (Cx + D)/(x^2 + 1)^2 = ((Ax + B)(x^2 + 1) + (Cx + D))/(x^2 + 1)^2 = (Ax^3 + Ax + Bx^2 + B + Cx + D)/(x^2 + 1)^2

x^3: A = 5
x^2: B = -3
x: A + C = 7; C = 2
1: B + D = -3; D = 0

∫(5x^3 - 3x^2 + 7x - 3)/(x^2 + 1)^2 dx = ∫(5x - 3)/(x^2 + 1) dx + ∫2x/(x^2 + 1)^2 dx = 5∫x/(x^2 + 1) dx - 3∫dx/(x^2 + 1) dx + ∫2x/(x^2 + 1)^2 dx = 5/2 ln|x^2 + 1| -3arctanx - 1/(x^2 + 1) + C



14) ∫(2x^4 - 2x^3 + 6x^2 - 5x + 1)/(x^3 - x^2 + x - 1) dx = (after long division) ∫(2x + (4x^2 - 3x + 1)/(x^2 + 1)(x - 1)) dx

Consider (4x^2 - 3x + 1)/(x^2 + 1)(x - 1) = (Ax+ B)/(x^2 + 1) + C/(x - 1) = ((Ax+ B)(x - 1) + C(x^2 + 1))/(x^2 + 1)(x - 1) = (Ax^2 - Ax + Bx - B) + C(x^2 + 1))/(x^2 + 1)(x - 1)

x^2: A + C = 4; A = 4 - C
x: -A + B = -3; B = (4 - C) - 3 = 1 - C
1: -B + C = 1; -(1 - C) + C = 1; C = 1, A = 3, B = 0

∫(2x + (4x^2 - 3x + 1)/(x^2 + 1)(x - 1)) dx = ∫2x dx + ∫3x/(x^2 + 1) dx + ∫dx/(x - 1) = x^2 + 3/2 ln|x^2 + 1| + ln|x - 1| + C



15) ∫dx/(x^2 - c^2) = ∫dx/(x - c)(x + c)

Consider 1/(x - c)(x + c) = A/(x - c) + B/(x + c) = (A(x + c) + B(x - c))/(x - c)(x + c)

x: A + B = 0; A = -B
1: Ac - Bc = 1; B= -1/2c, A = 1/2c

∫dx/(x - c)(x + c) = 1/2c∫dx/(x - c) - 1/2c∫dx/(x + c) = 1/2c ln|(x - c)/(x + c)| + C

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 1:30

Can you jump to the volume and area thing?

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 1:41

Ok, I'll be back tomorrow.
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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 1:43

What? Not again! How dare you sleep! 08

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 1:44

Hey, do you want me to help? =D

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 1:46

No. =P

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 1:47

Oh, how welcoming. I wasn't going to help you anyway. XD

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