Question

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

Answer 1

The integral of that you mean?

Answer 2

Yes, sorry I wasn't clear

Answer 3

Definite Integral

Answer 4

let u=6t+1

Answer 5

Use a u substation as myininaya has just suggested ;p

Answer 6

substitution I mean.

Answer 7

I have to use the Fundamental Theorem of Calculus

Answer 8

we will

Answer 9

Ok

Answer 10

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

Answer 11

haha

Answer 12

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

Answer 13

or maybe impossible

Answer 14

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

Answer 15

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

Answer 16

do you know substation?

Answer 17

substitution ;p

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

6

Answer 23

right do du=....

Answer 24

6 dt?

Answer 25

right

Answer 26

Then I hit both sides with integrals, correct?

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

and dt = 1/6 du

so \[\int_0^4 \frac{1}{6t+1}dt = \int_1^{25} \frac{1}{u}(\frac{1}{6} du)\]

Answer 34

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

Answer 35

when i say x i meant t lol

Answer 36

Hmmmm...

Answer 37

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
\[\int_0^4 \frac{1}{6t+1} dt = \int_1^{25} \frac{1}{u}(\frac{1}{6}du) \]
\[= \frac{1}{6} \int_1^{25} \frac{1}{u}du\]

Answer 38

Make more sense?

Answer 39

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

Answer 40

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.

Answer 41

Ohhh, okay, I see

Answer 42

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

Answer 43

i just posted it

Answer 44

1 over that right?

Answer 45

Yes

Answer 46

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

Answer 47

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.

Answer 48

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

Answer 49

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?

Answer 50

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

Answer 51

I'm not sure, that's all I was given and I'm not too sharp in this subject of integrals yet, sorry.

Answer 52

polpak come back? lol

Answer 53

Haha

Answer 54

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

Answer 55

Thank you so much! Please! lol

Answer 56

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

Answer 57

I'm so stressed out. Lol

Answer 58

Where did the 5 come from?

Answer 59

which 5?

Answer 60

Oh duh, I was looking at it too hard.

Answer 61

so you are teaching yourself?

Answer 62

Yes

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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.

Answer 67

right

Answer 68

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

Answer 69

Sorry , what's up?

Answer 70

Oh, 2000 is t=0

Answer 71

ok cool lol

Answer 72

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

Answer 73

t=10 would be 2010

Answer 74

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

Answer 75

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

Answer 76

np =)

Answer 77

Thank you polpak!

Answer 78

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

Answer 79

What did you get after evaluation?

Answer 80

I got 40 million people