Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno

Go back

Topic review (newest first)

2006-03-25 17:27:05


(yu)'=y'u+yu'   (yu)'dx= (y'u+yu')dx= udy+ydu

∫udy= ∫(yu)'dx -∫ydu =yu-∫ydu

use this to solve Q2

2006-03-25 15:38:32

∫ e^(tx) / β  e^(-x/β)  dx
do you mean it by ∫etx/βe-x/β dx ?
or ∫etx/(β e-x/β) dx?

2006-03-24 13:04:05

i'm not good at integration and i'm not sure whether i did it correct or not

1st tech

since you wanna find its integral, or antiderivative, F(x) is one of the answers if its derivative is truely the one whose derivative you wanna find

2nd tech

F(x)+Cs  share the same derivative with F(x), F(x)+C is also the antiderivative of F'(x)

3rd tech

given another one function satisfying G'(x)=F'(x) G(x) must be of the form F(x)+C

for (G(x)-F(x))'=G'(x)-F'(x)=0 and only C'=0

Conclusion: once you've found a function F(x) whose derivative is the one you wanna integrate, F(x)+c represents any its antiderivative

4th tech

(Cf(x))'=C(f(x))' or=Cf'(x) where C is a constant

5th tech

f'(x)=dy/dx x is any variable defined in domain

hence x itself could be a function too, as long as its value vary in the domain

x=g(z) g'(z)=dx/dz

it makes sense that


hence dy/dz= (f(g(z)))' of z=dy/dx dx/dz= f'(x) g'(z)=f'(g(z)) g'(z)

substitude x for z in the formula above

using tech 3 tech4 and tech5, do you understand how to solve problem 1 now?

2006-03-24 02:27:31

thanks for solving the problem.
Could you please tell me what techniques u used in solving those problems.
Could you please help me doing the third one..plzzzzzzzzzzzzzzzz
i would be more than gradeful

2006-03-24 01:58:33

Suggest you to get a Derive software

i am bored with integration solving, haha

1) [e^(-x/β)]'=e^(-x/β) (-x/β)'= - 1/β e^(-x/β)

∫ 1/β e^(-x/β)  dx = -e^(-x/β)+C

2) ∫ x/β e^(-x/β)  dx =  ∫xd(-e^(-x/β)) = -∫xd(e^(-x/β))= -xe^(-x/β)+∫e^(-x/β)dx =  -xe^(-x/β) - βe^(-x/β)+ C

i am tired...

2006-03-23 22:40:40

Hello guyzz..
I wonder if somebody can evaluate the following integrals

1.     ∫ 1/β e^(-x/β)  dx

2.     ∫ x/β e^(-x/β)  dx

3.     ∫ e^(tx) / β  e^(-x/β)  dx

lets see who of you guyzz can solve it.

James Michael

Board footer

Powered by FluxBB