# Thread: Rate of oil production..

1. ## 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?!

2. 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 $\displaystyle \int_0^5R(t)~dt$

3. 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 !!

4. 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, $\displaystyle \int cx^n~dx = \frac {cx^{n + 1}}{n + 1} + C$

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

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

where $\displaystyle c$ is a constant

5. 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

6. 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

7. oooops !! i meant, 39.87 for that one! EXCELLENT! thanks you soo much!!!