Definite integral prob (1 Viewer)

[flyingKangaroo]

Hey, was hoping someoen could explain how to do this. I know it should equal 1 but whenever I solve I always end up with 5/4 or something.

(5x(2-x)^3)/8 for the domain [0,2]

 

[Drongoski]

Hey, was hoping someoen could explain how to do this. I know it should equal 1 but whenever I solve I always end up with 5/4 or something.

(5x(2-x)^3)/8 for the domain [0,2]

Don't understand question. Plz clarify.

 

[GUSSSSSSSSSSSSS]

u = 2-x --> x = 2-u
du/dx = -1
dx = -du
when x = 2, u =0
when x = 0, u = 2
therfore
5/8 (2-u)(u)^3 .-du
=-5/8(2u^3 - u^4)du
= 5/8 [(u^4)/2 - (u^5)/5] 2 -> 0 (LOL dunno how to write that XD)
=5/8(8 - 32/5 - (0 - 0))
=5/8((40-32)/5)
=1


XDD you get?

 

[EvoRevolution]

to make it look abit nicer

 

[Drongoski]

Now I get it; I thought it was

 

[GUSSSSSSSSSSSSS]

GODAMIT lol my solutions always look so messy (thats why i hate doin integration problems on here)
CANT use that latex thingy, im too technologically illiterate (however i guess i got nothin to complain about when drongoski is around =P )

 

[Drongoski]

GODAMIT lol my solutions always look so messy (thats why i hate doin integration problems on here)
CANT use that latex thingy, im too technologically illiterate (however i guess i got nothin to complain about when drongoski is around =P )

Hey guss .. ss

u shld have seen the way I was plodding away with my one-finger typing !! Took me 20 mins. I know u put yrs up already; but I thought my sustitutionless soln may be of some interest.

 

[GUSSSSSSSSSSSSS]

Hey guss .. ss

u shld have seen the way I awas plodding away with my one-finger typing !! Took me 20 mins. I know u put yrs up already; but I thought my sustitutionless soln may be of some interest.

HAHAHA well, im DEFINITELY not as bad as u then with technology, but im not that great =S

yess i read your solution, i never thought of doing it that way, i guess i shud try it that way next couple of integration q's that i do, see if it is quicker!
THANKS!!

 

[flyingKangaroo]

hey thanks for that.
i understand most of it, though i don't get the substitution of (2-x) for u and the 'when x=2, u=0' etc. part...could you please explain that?

thanksss

 

[oly1991]

you need to change the x-values to u-values because you are now intergrating u-values because you made u=2-x.

 

[GUSSSSSSSSSSSSS]

well 2-x was substituted by u in an attempt to simplify the integral, which it did

the values have to be changed because now we are integrating with respect to u (we have a du, rather than a dx)
thus the boundaries of the integral have to be changed to be values that correspond with u

make sure you do that EVERYTIME u evaluate an integral which involved a substitution