can someone explain this (1 Viewer)

[xeriphic]

when i integrate 1/2x

int 1/2x dx = 1/2 int 2/2x dx

= 1/2 ln 2x + c

but if i had simplified before where

int 1/2x dx = 1/2 int 2/2x dx = 1/2 int 1/x dx

= 1/2 ln x + c

are both solutions valid, but the answer are different, can someone clarify the correct way to do it

 

[xeriphic]

also how do i do

1) int (4x+3)/(2x+1) dx

2) int (2x+5)/(x+4) dx

3) the region under the curve y = sqrt 2x/(x^2+1) and bounded by the lines x=0 and x=1 is rotated about the x-axis. Find the volume of the revolution

thanks

 

[untamedanimal]

Originally posted by xeriphic
when i integrate 1/2x

int 1/2x dx = 1/2 int 2/2x dx

= 1/2 ln 2x + c

but if i had simplified before where

int 1/2x dx = 1/2 int 2/2x dx = 1/2 int 1/x dx

= 1/2 ln x + c

are both solutions valid, but the answer are different, can someone clarify the correct way to do it

when you get int 1/2x, the 1/2 should be taken out of the the integration, just like you would take 3 out of the integration in the int of 3x.

So ite ends up being 1/2 int 1/x = 1/2 ln x + c

 

[xeriphic]

what about my other solution isn't it valid as well since

int 1/2x dx = 1/2 int 2/2x dx

= 1/2 ln 2x + c

not sure where this one would be wrong

 

[:: ck ::]

i think when u integrate u find the simplest expression in the integrand.....

Originally posted by xeriphic
also how do i do

1) int (4x+3)/(2x+1) dx

2) int (2x+5)/(x+4) dx

3) the region under the curve y = sqrt 2x/(x^2+1) and bounded by the lines x=0 and x=1 is rotated about the x-axis. Find the volume of the revolution

thanks

1 and 2 are just basic manipulation

1) integral (4x+3)/(2x+1)dx
= integral [(4x+2)/(2x+1) + 1/(2x+1) ] dx
= 2x + 1/2 integral [2/(2x+1)]
= 2x + (1/2)ln|2x+1| + c

2) int (2x+5)/(x+4) dx
= integral (2x+8) / (x+4) - 3/(x+4) dx
= 2x - 3ln|x+4| + C

3) mmm dunno wot else to say but the basic steps...

square ur expression for y

then use V = pi integral y² dx

 

[Winston]

Originally posted by :: ryan.cck ::
i think when u integrate u find the simplest expression in the integrand.....



1 and 2 are just basic manipulation

1) integral (4x+3)/(2x+1)dx
= integral [(4x+2)/(2x+1) + 1/(2x+1) ] dx
= 2x + 1/2 integral [2/(2x+1)]
= 2x + (1/2)ln|2x+1| + c

Heh i remember that question it should be frm the J.B.Couchman textbook. book 2.

Anyways... what your aim is in that type of question is
is to modify the numerator so that it would enable to be cancelled out with the denominator

see how 4x +3, how ryan turned it into 4x +2 + 1,

so basically you just need to manipulate the numerator in the sense that

it can be simplified with the denominator and also the numerator will still hold true to what it originally was.

 

[xeriphic]

lol Winston, indeed it is from that text book, pretty helpful i think, i think i understand them now

thanks guys for the help

 

[Winston]

Originally posted by xeriphic
lol Winston, indeed it is from that text book, pretty helpful i think, i think i understand them now

thanks guys for the help

Haha yeah i miss that book, i think it was the best and most succinct ...

What school do you go to?

 

[xeriphic]

i'm going to khs, i think most school in this area uses this text book

by the way can someone help me with this question

1) int ((x^2)(e^2x^2) - 1) / x dx

2) show that (x+7)/(x-1) = 1+ 8/(x-1)

thanks

 

[Xayma]

2) x+7=x-1+8
.: (x+7)/(x-1)
=(x-1)/(x-1)+8/(x-1)
=1+8/(x-1)

 

[xeriphic]

thanks, i understand it though do you have any hints on answering these kind of questions

 

[Heinz]

Originally posted by xeriphic


1) int ((x^2)(e^2x^2) - 1) / x dx

thanks

Im trying to think of a 2unit method but in the meantime...

int [(x²)(e^2x² - 1)] / x dx
= int x((e^2x²) - 1) dx

let u = 2x² :. du/dx = 4x

int x((e^2x²) - 1) dx
= [1/4] int e^u - 1 du
= [1/4][e^u - u] + c
= [1/4][e^2x²- 2x²] + c

 

[xeriphic]

Originally posted by Heinz
int x((e^2x²) - 1) dx
= [1/4] int e^u - 1 du
= [1/4][e^u - u] + c
= [1/4][e^2x²- 2x²] + c

i'm lost in this part

 

[Heinz]

Originally posted by xeriphic
i'm lost in this part

let u = 2x² :. du/dx = 4x :.dx = du/4x

int x((e^2x²) - 1) dx
wherever you see 2x² put "u" and wherever you see dx put "du/4x"
= int x(e^u -1)*du/4x
= int [1/4][e^u - 1] du
= [1/4] int e^u - 1 du
= [1/4][e^u - u] + c
= [1/4][e^2x²- 2x²] + c

Does that help? This is using integration by substitution (3unit)

Edit: couple of typos

 

[xeriphic]

actually the answer was (1/4)(e^2x^2 - ln x) + c , so ln x would be from 1/x some how

 

[CrashOveride]

I think Milhouse has misquoted your original question, hence the discrepancy in the answer.

If you split it up via partial fractions, you'll have x²e^2x² / x - 1/x

That's where the ln x comes from.
If you just do what Milhouse did via substitution method, you should get 1/4e^2x² - lnx + C

 

[Heinz]

Sorry, my mistake. I did read it incorrectly (damn parentheses ). The process is the same though. Just change it to suit the question and you should be fine.

 

[CM_Tutor]

Originally posted by xeriphic
when i integrate 1/2x

int 1/2x dx = 1/2 int 2/2x dx

= 1/2 ln 2x + c

but if i had simplified before where

int 1/2x dx = 1/2 int 2/2x dx = 1/2 int 1/x dx

= 1/2 ln x + c

are both solutions valid, but the answer are different, can someone clarify the correct way to do it

Both solutions are valid - in fact, both solutions are the same. The (1 / 2)ln x + C solution is neater, but both are true.

 

[:: ck ::]

if it was a definite integral wouldnt the answer differ?!

 

[Xayma]

No.

Eg. if the end points are 1 to 3.
1/2(ln3-ln1)=0.549
1/2(ln6-ln2)=0.549

Because you cant have ln 0, any increase in the first one is cancelled out by the increase in the second.