# Rate of oil production..

• March 30th 2008, 09:28 PM
melanie
Rate of oil production..
Here's a problem. Not sure if I should just sub in or firstly integrate and then sub in. Please help..

an oil company found that the rate of oil production from an oil well is given by a function of time t, in years:

R(t) = [ 60t/t^2 +9 ] + 35

Approx. how many barrels of oil will be produced in the first 5 years of production?!
• March 30th 2008, 09:35 PM
Jhevon
Quote:

Originally Posted by melanie
Here's a problem. Not sure if I should just sub in or firstly integrate and then sub in. Please help..

an oil company found that the rate of oil production from an oil well is given by a function of time t, in years:

R(t) = [ 60t/t^2 +9 ] + 35

Approx. how many barrels of oil will be produced in the first 5 years of production?!

you want $\int_0^5R(t)~dt$
• March 30th 2008, 09:59 PM
melanie
Thank you very much for that tip!!!

But another question is . What happens to the constant (35) when integrating. DO i just ignore it?!

So, (im not sure how to use the integration sign on the computer)

'Integration sign (0-5 years) = |' [60t/t^2+9] +53 dt
= 60/2 | [1/t^2+9] 2t dt

let u = t^2 +9
then du = 2t dt (?)

= 60/2 | 1/u du

= 60/2 ln |u| + c

= 30 ln (t^2 +9 )

= 30 ln 34 - 30 ln 9
= 105.8 - 65.92
= approx. 40 ?

I really appreciate your help !!
• March 30th 2008, 10:01 PM
Jhevon
Quote:

Originally Posted by melanie
Thank you very much for that tip!!!

But another question is . What happens to the constant (35) when integrating. DO i just ignore it?!

So, (im not sure how to use the integration sign on the computer)

'Integration sign (0-5 years) = |' [60t/t^2+9] +53 dt
= 60/2 | [1/t^2+9] 2t dt

let u = t^2 +9
then du = 2t dt (?)

= 60/2 | 1/u du

= 60/2 ln |u| + c

= 30 ln (t^2 +9 )

= 30 ln 34 - 30 ln 9
= 105.8 - 65.92
= approx. 40 ?

I really appreciate your help !!

the rule is, $\int cx^n~dx = \frac {cx^{n + 1}}{n + 1} + C$

note that if you have a constant, the $n$ above is zero. since $c = cx^0$

thus, $\int c~dx = cx + C$

where $c$ is a constant
• March 30th 2008, 10:18 PM
melanie
Hmm.. ok.

So, I integrate 60t/t^2+9 AND 35 separately.

I get:

30 ln 34 - 30 ln 9 = 215 [ for the first one ]

35(5) - 35(0) = 175 [ from integrating the constant ]

and I add them both together:
= 215.

Is that right ?! :S
• March 30th 2008, 10:21 PM
Jhevon
Quote:

Originally Posted by melanie
Hmm.. ok.

So, I integrate 60t/t^2+9 AND 35 separately.

I get:

30 ln 34 - 30 ln 9 = 215 [ for the first one ]

35(5) - 35(0) = 175 [ from integrating the constant ]

and I add them both together:
= 215.

Is that right ?! :S

your final answer is right. but 30ln(34) - 30ln(9) is not 215

of course, the answer is approximate, so you should write the squiggly equal sign
• March 30th 2008, 10:23 PM
melanie
oooops !! i meant, 39.87 for that one! EXCELLENT! thanks you soo much!!! (Rofl)