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anonymous
 5 years ago
1/(6t+1)^1/2 dt evaluated from 0 to 4
anonymous
 5 years ago
1/(6t+1)^1/2 dt evaluated from 0 to 4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The integral of that you mean?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, sorry I wasn't clear

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use a u substation as myininaya has just suggested ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have to use the Fundamental Theorem of Calculus

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0there is no other way unless you used riemann sums but lets not mention him anymore

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0using riemann sums can also be very difficult to evaluate the limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well I'm teaching myself Integrals so I'm not quite that far yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm just having trouble finding where to start with these functions.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0do you know substation?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0stupid thingy i hit whatever popped up because i hate spelling substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to learn u substitutions to do these they allow you to undo the chain rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, So after the substitution, What's my next step?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0if u=6t+1 then du/dt=.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then I hit both sides with integrals, correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. Then you substitute 6t +1 as u, and your dt as 1/6 du.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And take the integral with respect to u (though you also have to change your limits)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, ok, that's where I get lost.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0polpak divided both sides by 6 of that equation du=6dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 6t + 1 = u the 1/(6t+1) = 1/u

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0and dt = 1/6 du so \[\int_0^4 \frac{1}{6t+1}dt = \int_1^{25} \frac{1}{u}(\frac{1}{6} du)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since 1/6 is just a multiplicative constant you can pull it out front.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0when i say x i meant t lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, 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
 5 years ago
Best ResponseYou've already chosen the best response.0The only thing I'm confused about is where \[\int\limits_{1}^{25}\] comes from

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0See 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
 5 years ago
Best ResponseYou've already chosen the best response.0But, the original equation it's \[\sqrt{6t+1}\] What does that change in your answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow, thanks for that effort in explaining, that helps so much!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, thank you both, very much appreciated guys, I have another question as well if you want to answer it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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 20002010?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0for 2000, t=0 is that right? t=1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure, that's all I was given and I'm not too sharp in this subject of integrals yet, sorry.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0polpak come back? lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we can work on integrating P(t) together let me look at it, and I will scan it in k?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you so much! Please! lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0hey so we have [e^(0.2t)]/4 + e^(0.2t) ={5e^(0.2t)}/4?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm so stressed out. Lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where did the 5 come from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh duh, I was looking at it too hard.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so you are teaching yourself?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0thats 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so how do you determine what limits to use?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0well it wants to know the average population from 2000 to 2010 so from there we can figure out the limits

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alrighty, so what difference would it make to use either one of the limits?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"(t) years after the year 2000" so if t = 0 the year is 2000.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0right i was not sure at first if it was 1 to 11 or 0 to 10 thanks polpak

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so you do [25e^(0.2*10)]/4  [25e^(0*10)]/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What did you get after evaluation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got 40 million people
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