anonymous
  • anonymous
1/(6t+1)^1/2 dt evaluated from 0 to 4
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
The integral of that you mean?
anonymous
  • anonymous
Yes, sorry I wasn't clear
anonymous
  • anonymous
Definite Integral

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myininaya
  • myininaya
let u=6t+1
anonymous
  • anonymous
Use a u substation as myininaya has just suggested ;p
anonymous
  • anonymous
substitution I mean.
anonymous
  • anonymous
I have to use the Fundamental Theorem of Calculus
myininaya
  • myininaya
we will
anonymous
  • anonymous
Ok
myininaya
  • myininaya
there is no other way unless you used riemann sums but lets not mention him anymore
anonymous
  • anonymous
haha
myininaya
  • myininaya
using riemann sums can also be very difficult to evaluate the limit
myininaya
  • myininaya
or maybe impossible
anonymous
  • anonymous
Well I'm teaching myself Integrals so I'm not quite that far yet.
anonymous
  • anonymous
I'm just having trouble finding where to start with these functions.
myininaya
  • myininaya
do you know substation?
anonymous
  • anonymous
substitution ;p
myininaya
  • myininaya
stupid thingy i hit whatever popped up because i hate spelling substitution
anonymous
  • anonymous
You have to learn u substitutions to do these they allow you to undo the chain rule.
anonymous
  • anonymous
Ok, So after the substitution, What's my next step?
myininaya
  • myininaya
if u=6t+1 then du/dt=.....
anonymous
  • anonymous
6
myininaya
  • myininaya
right do du=....
anonymous
  • anonymous
6 dt?
myininaya
  • myininaya
right
anonymous
  • anonymous
Then I hit both sides with integrals, correct?
anonymous
  • anonymous
No. Then you substitute 6t +1 as u, and your dt as 1/6 du.
anonymous
  • anonymous
And take the integral with respect to u (though you also have to change your limits)
anonymous
  • anonymous
Oh, ok, that's where I get lost.
myininaya
  • myininaya
polpak divided both sides by 6 of that equation du=6dt
anonymous
  • anonymous
Well 6t + 1 = u the 1/(6t+1) = 1/u
myininaya
  • myininaya
you dont have to change the limits you can just leave it blank and come back after finding the antiderivative in terms of x in using the limits that we started with
anonymous
  • anonymous
and dt = 1/6 du so \[\int_0^4 \frac{1}{6t+1}dt = \int_1^{25} \frac{1}{u}(\frac{1}{6} du)\]
anonymous
  • anonymous
Since 1/6 is just a multiplicative constant you can pull it out front.
myininaya
  • myininaya
when i say x i meant t lol
anonymous
  • anonymous
Hmmmm...
anonymous
  • anonymous
Ok, let's try this one step at a time. Let u = 6t +1 \[\implies du = 6dt \implies dt = \frac{1}{6}du\] \[t = 0 \implies u = 0 + 1 = 1\] \[t = 4 \implies u = 6(4) + 1 = 25\] Therefore \[\int_0^4 \frac{1}{6t+1} dt = \int_1^{25} \frac{1}{u}(\frac{1}{6}du) \] \[= \frac{1}{6} \int_1^{25} \frac{1}{u}du\]
anonymous
  • anonymous
Make more sense?
anonymous
  • anonymous
The only thing I'm confused about is where \[\int\limits_{1}^{25}\] comes from
anonymous
  • anonymous
See the t=0 and t=4 lines? If we are evaluating t from 0 to 4, that means u goes from 1 to 25.
anonymous
  • anonymous
Ohhh, okay, I see
myininaya
  • myininaya
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anonymous
  • anonymous
But, the original equation it's \[\sqrt{6t+1}\] What does that change in your answer?
myininaya
  • myininaya
i just posted it
myininaya
  • myininaya
1 over that right?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Wow, thanks for that effort in explaining, that helps so much!
anonymous
  • anonymous
Oh, sorry didn't see the exponent. It changes things some in that it's \(1/u^2\) instead of 1/u, but that just changes the end result, not too much in what I posted.
anonymous
  • anonymous
Ok, thank you both, very much appreciated guys, I have another question as well if you want to answer it.
myininaya
  • myininaya
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anonymous
  • anonymous
The population of a certain community (t) years after the year 2000 is given by \[P(t)= e ^{0.2t}\div4+e ^{0.2t} million people\] what was the average population from 2000-2010?
myininaya
  • myininaya
for 2000, t=0 is that right? t=1?
anonymous
  • anonymous
I'm not sure, that's all I was given and I'm not too sharp in this subject of integrals yet, sorry.
myininaya
  • myininaya
polpak come back? lol
anonymous
  • anonymous
Haha
myininaya
  • myininaya
we can work on integrating P(t) together let me look at it, and I will scan it in k?
anonymous
  • anonymous
Thank you so much! Please! lol
myininaya
  • myininaya
hey so we have [e^(0.2t)]/4 + e^(0.2t) ={5e^(0.2t)}/4?
anonymous
  • anonymous
I'm so stressed out. Lol
myininaya
  • myininaya
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anonymous
  • anonymous
Where did the 5 come from?
myininaya
  • myininaya
which 5?
anonymous
  • anonymous
Oh duh, I was looking at it too hard.
myininaya
  • myininaya
so you are teaching yourself?
anonymous
  • anonymous
Yes
myininaya
  • myininaya
thats cool. we need to figure out what limits to use? its ether 1 to 11 or 0 to 10 depending if we start 2000 with t=1 or t=0 i will get polpak and see if he knows
anonymous
  • anonymous
Ok, so how do you determine what limits to use?
myininaya
  • myininaya
well it wants to know the average population from 2000 to 2010 so from there we can figure out the limits
anonymous
  • anonymous
Ok, so the t=1 or t=0 would represent the year 2000, as the 10/11 would represent the 10 year time frame from 2000.
myininaya
  • myininaya
right
anonymous
  • anonymous
Alrighty, so what difference would it make to use either one of the limits?
anonymous
  • anonymous
Sorry , what's up?
anonymous
  • anonymous
Oh, 2000 is t=0
myininaya
  • myininaya
ok cool lol
anonymous
  • anonymous
"(t) years after the year 2000" so if t = 0 the year is 2000.
anonymous
  • anonymous
t=10 would be 2010
myininaya
  • myininaya
ok so 2010 would be t=10 so we have our upper and lower limit just use the fundamental thm of calculus on the antiderivative we found
myininaya
  • myininaya
right i was not sure at first if it was 1 to 11 or 0 to 10 thanks polpak
anonymous
  • anonymous
np =)
anonymous
  • anonymous
Thank you polpak!
myininaya
  • myininaya
so you do [25e^(0.2*10)]/4 - [25e^(0*10)]/4
anonymous
  • anonymous
What did you get after evaluation?
anonymous
  • anonymous
I got 40 million people

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