1/(6t+1)^1/2 dt evaluated from 0 to 4

- anonymous

1/(6t+1)^1/2 dt evaluated from 0 to 4

- schrodinger

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- anonymous

The integral of that you mean?

- anonymous

Yes, sorry I wasn't clear

- anonymous

Definite Integral

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## More answers

- myininaya

let u=6t+1

- anonymous

Use a u substation as myininaya has just suggested ;p

- anonymous

substitution I mean.

- anonymous

I have to use the Fundamental Theorem of Calculus

- myininaya

we will

- anonymous

Ok

- myininaya

there is no other way unless you used riemann sums but lets not mention him anymore

- anonymous

haha

- myininaya

using riemann sums can also be very difficult to evaluate the limit

- myininaya

or maybe impossible

- anonymous

Well I'm teaching myself Integrals so I'm not quite that far yet.

- anonymous

I'm just having trouble finding where to start with these functions.

- myininaya

do you know substation?

- anonymous

substitution ;p

- myininaya

stupid thingy i hit whatever popped up because i hate spelling substitution

- anonymous

You have to learn u substitutions to do these they allow you to undo the chain rule.

- anonymous

Ok, So after the substitution, What's my next step?

- myininaya

if u=6t+1
then du/dt=.....

- anonymous

6

- myininaya

right do du=....

- anonymous

6 dt?

- myininaya

right

- anonymous

Then I hit both sides with integrals, correct?

- anonymous

No. Then you substitute 6t +1 as u, and your dt as 1/6 du.

- anonymous

And take the integral with respect to u (though you also have to change your limits)

- anonymous

Oh, ok, that's where I get lost.

- myininaya

polpak divided both sides by 6 of that equation du=6dt

- anonymous

Well 6t + 1 = u the 1/(6t+1) = 1/u

- 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

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

Since 1/6 is just a multiplicative constant you can pull it out front.

- myininaya

when i say x i meant t lol

- anonymous

Hmmmm...

- 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

Make more sense?

- anonymous

The only thing I'm confused about is where \[\int\limits_{1}^{25}\] comes from

- 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

Ohhh, okay, I see

- myininaya

##### 1 Attachment

- anonymous

But, the original equation it's \[\sqrt{6t+1}\]
What does that change in your answer?

- myininaya

i just posted it

- myininaya

1 over that right?

- anonymous

Yes

- anonymous

Wow, thanks for that effort in explaining, that helps so much!

- 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

Ok, thank you both, very much appreciated guys, I have another question as well if you want to answer it.

- myininaya

##### 1 Attachment

- 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

for 2000, t=0 is that right? t=1?

- 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

polpak come back? lol

- anonymous

Haha

- myininaya

we can work on integrating P(t) together let me look at it, and I will scan it in k?

- anonymous

Thank you so much! Please! lol

- myininaya

hey so we have [e^(0.2t)]/4 + e^(0.2t) ={5e^(0.2t)}/4?

- anonymous

I'm so stressed out. Lol

- myininaya

##### 1 Attachment

- anonymous

Where did the 5 come from?

- myininaya

which 5?

- anonymous

Oh duh, I was looking at it too hard.

- myininaya

so you are teaching yourself?

- anonymous

Yes

- 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

Ok, so how do you determine what limits to use?

- myininaya

well it wants to know the average population from 2000 to 2010 so from there we can figure out the limits

- 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

right

- anonymous

Alrighty, so what difference would it make to use either one of the limits?

- anonymous

Sorry , what's up?

- anonymous

Oh, 2000 is t=0

- myininaya

ok cool lol

- anonymous

"(t) years after the year 2000" so if t = 0 the year is 2000.

- anonymous

t=10 would be 2010

- 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

right i was not sure at first
if it was 1 to 11
or 0 to 10 thanks polpak

- anonymous

np =)

- anonymous

Thank you polpak!

- myininaya

so you do [25e^(0.2*10)]/4 - [25e^(0*10)]/4

- anonymous

What did you get after evaluation?

- anonymous

I got 40 million people

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