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integrate
- Thread starter Mel Mitch
- Start date
Hello, Mel Mitch!
I have no idea what you were trying to do.
This integral requires a substitution . . .
\(\displaystyle \text{Let: }\:u \:=\:\sqrt{x+1} \quad\Rightarrow\quad x \:=\:u^2-1 \quad\Rightarrow\quad dx \:=\:2u\,du\)
\(\displaystyle \text{Substitute: } \;\int^3_2\frac{3(u^2-1) + 2}{u}\,2u\,du \;\;=\;\;2\int^3_2(3u^2-1)\,du \;=\;2(u^3 - u)\,\bigg]^3_2\)
\(\displaystyle \text{Evaluate: }\;2(3^3-3) - 2(2^3-2) \;\;=\;\;2(24) - 2(6) \;\;=\;\;48 - 12 \;\;=\;\;\boxed{36}\)
I have no idea what you were trying to do.
This integral requires a substitution . . .
\(\displaystyle \text{Evaluate: }\;\int^8_3\frac{3x+2}{\sqrt{x+1}}\,dx\)
\(\displaystyle \text{Let: }\:u \:=\:\sqrt{x+1} \quad\Rightarrow\quad x \:=\:u^2-1 \quad\Rightarrow\quad dx \:=\:2u\,du\)
\(\displaystyle \text{Substitute: } \;\int^3_2\frac{3(u^2-1) + 2}{u}\,2u\,du \;\;=\;\;2\int^3_2(3u^2-1)\,du \;=\;2(u^3 - u)\,\bigg]^3_2\)
\(\displaystyle \text{Evaluate: }\;2(3^3-3) - 2(2^3-2) \;\;=\;\;2(24) - 2(6) \;\;=\;\;48 - 12 \;\;=\;\;\boxed{36}\)
D
Deleted member 4993
Guest
Mel Mitch said:Thanks....i follow u ....i don't understand why u changed and used the values x=3 and x=2 ....instead of 8 and 3....
When x = 3 ? how much is u?
When x = 8 ? how much is u?
Previously you had said you were doing CalcIII problems - I hope you plan to review "integration" thoroughly.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
Right on Subhotosh Khan, I think we are being "had" by mel mitch as if he is doing calculus, all this stuff should be second nature to him.
i think u r missing the point if u arent willing the help me..... that's find......i dont think u understand when i really type "i dont understand' ....but if u like to find out ....Why i'm on this site...or any other personal info...u could send me a personal message......not too personal cause the internet got weirdos.
but to SUB
I replaced u=(1+x)^1/2 and substituted its value and got 36.... ok
note if this was something i did understand....i would basically wouldn't need this website....as i said i'm studying for a test and please guide me.....
my lec behaves just like u GUYS (males) that's why so many aren't doing it any longer....that is....like u'r suppose to know everything in maths once ur doing it.....but its ok.....
but to SUB
I replaced u=(1+x)^1/2 and substituted its value and got 36.... ok
note if this was something i did understand....i would basically wouldn't need this website....as i said i'm studying for a test and please guide me.....
my lec behaves just like u GUYS (males) that's why so many aren't doing it any longer....that is....like u'r suppose to know everything in maths once ur doing it.....but its ok.....
D
Deleted member 4993
Guest
but to SUB
I replaced u=(1+x)^1/2 and substituted its value and got 36.... ok <<< Can you please show me the steps - how you got that?
I think you do not remember - have not practiced enough - what you did before (as a pre-req to this class). This type of substituition (evaluation of function) is taught in "begining Algebra". In a Calc III class, instructor has the "right" to assume that you know that part.
Ok
After placing 'u'
where x=u^2-1 and u=(x+1)^1/2
integrating (3x+2)/(x+1)^1/2 2udu
int 2 [3(u^2-1)+2]/u
int 2[(3u^2-1) x u^-1
2[u^3-u]
2[u(u^2-u)] when x=8, x=3
replacing u with x
2[(x+1)(x+1-1)]
2.(x+1)^1/2. x
(2.sqrt9.8)-(2.sqrt4.3)
48-12=36
It is an upgrading course that my country suggest we do as a teachers teaching between grades 4-11......we had a choice between English language and Mathematics.....and since English is something i wouldn't enjoy and much hard i believe....then their was no question that Maths was it.......
I do agree with u.... with the practice....something i feel like i'm reading a some strange language
so far we did: functions, vectors,coordinate geometry1,trig1,algebra1,matrices,permutations and combinations,series and binomial theorem, probability,cal1,sketching functions1,cal11,cal111
discussion for sketching function11 and trig11 starting saturday and the other week we finish off with coordinate geometry
After placing 'u'
where x=u^2-1 and u=(x+1)^1/2
integrating (3x+2)/(x+1)^1/2 2udu
int 2 [3(u^2-1)+2]/u
int 2[(3u^2-1) x u^-1
2[u^3-u]
2[u(u^2-u)] when x=8, x=3
replacing u with x
2[(x+1)(x+1-1)]
2.(x+1)^1/2. x
(2.sqrt9.8)-(2.sqrt4.3)
48-12=36
It is an upgrading course that my country suggest we do as a teachers teaching between grades 4-11......we had a choice between English language and Mathematics.....and since English is something i wouldn't enjoy and much hard i believe....then their was no question that Maths was it.......
I do agree with u.... with the practice....something i feel like i'm reading a some strange language
so far we did: functions, vectors,coordinate geometry1,trig1,algebra1,matrices,permutations and combinations,series and binomial theorem, probability,cal1,sketching functions1,cal11,cal111
discussion for sketching function11 and trig11 starting saturday and the other week we finish off with coordinate geometry