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You are not logged in. #1 20050911 23:20:48
Integrals: substitution methodhi everyone, plz don't laugh at me, i'm a math noob, i'm currently learning integrals and i got a problem. i dont understand how to solve this exercise: #2 20050911 23:55:42
Re: Integrals: substitution methodIf you are learning integration you are less n00b than you think...: ) #3 20050912 00:10:45
Re: Integrals: substitution method∫ [ (3x5)^(1/4) ]^3 dx = ∫ (3x5)^(3/4) dx #5 20050912 09:49:32
Re: Integrals: substitution methodok! I'm always here for integration problems : ) #6 20050913 02:33:28
Re: Integrals: substitution methodhey kyle...i got another one for you. i want u to solve this one, because i think that the solution in the book is not correct. the book was written by some of my teachers, they probably wrote the darn book in one night. so many mistakes. #7 20050914 06:42:44
Re: Integrals: substitution methodThe solution is correct. #8 20050914 06:53:57
Re: Integrals: substitution methodCode:let z = 25x dz —— = 5 dx dz = 5.dx ⌠ 3dx 3 ⌠ dx 3 1 ⌠ 5dx 3 ⌠ dz │——————————— = .│————————— =  —.—.│————————— = ——.│———— │ 4/5 4 │ 4/5 4 5 │ 4/5 20 │ 4/5 ⌡ 4(25x) ⌡(25x) ⌡(25x) ⌡z 3 let α =  —— and β = 4/5 to make things more clear. 20 ┌ ┐ │ β+1 │ ⌠ dz ⌠ 1 ⌠ β │z │ α β+1 α│——— = α│——— dz = α ⌡z dz = α│——————│ + C = ————.z + C │ β │ β │ β+1 │ β+1 ⌡ z ⌡ z └ ┘ β+1 = 1/5 α 3 ———— =  — β+1 4 putting everything together we get: 3 1/5  —(25x) + C 4 and that's it : ) Last edited by kylekatarn (20050914 06:55:56) #9 20050914 07:21:18
Re: Integrals: substitution methodExcellent! "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #10 20050914 07:49:35
Re: Integrals: substitution method
thanks:) #14 20060110 06:01:50
Re: Integrals: substitution methodFor future reference, if you want to ask for help, then create a new topic rather than posting on someone else's. Why did the vector cross the road? It wanted to be normal. 