# Using production and geological data, the management of an oil company estimates that oil will be...

## Question:

Using production and geological data, the management of an oil company estimates that oil will be pumped from a producing field at a rate given by

{eq}R(t) = \frac {100}{t+1} + 5; 0 \leq t \leq 20 {/eq}

where {eq}R(t) {/eq} is the rate of production (in thousands of barrel per year) t years after pumping begins.

Approximately, how many barrels of oil will the field produce during the first year of production? And from the end of the 10th year to the end of the 20th year of production?

## Definite Integrals:

Definite integrals are calculated via the Newton-Leibniz formula.

It states that the difference of the function values of the antiderivative of {eq}f(x) {/eq} at the endpoints of the interval {eq}\left[ a, b\right] {/eq} is calculated whenever we want to calculate its definite integral over this interval.

The given problem is an application of definite integrals.

The barrels of oil produced during the first year of production corresponds to the solution of the definite integral {eq}\displaystyle \int_{0}^{1} \left(\frac {100}{t+1} + 5\right)\, \mathrm{d}t {/eq}.

Meanwhile, the definite integral {eq}\displaystyle \int_{10}^{20} \left(\frac {100}{t+1} + 5\right)\, \mathrm{d}t {/eq} calculates the barrels of oil produced from the end of the 10th year to the end of the 20th year of production.

We must employ the Newton-Leibniz formula to both these definite integrals so we can attain the required solutions:

{eq}\begin{align*} \displaystyle \int_{0}^{1} \left(\frac {100}{t+1} + 5\right)\, \mathrm{d}t & = \left[100 \ln|t+1| +5t \right]_{0}^{1}\\ & =100 \ln|1+1| +5(1) -(100 \ln|0+1| +5(0) )&& \left[\mathrm{Newton-Leibniz \ formula }\right]\\ & =100 \ln(2) + 5\\ \implies \displaystyle \int_{0}^{1} \left(\frac {100}{t+1} + 5\right)\, \mathrm{d}t& \approx 74.315\\ \displaystyle \int_{10}^{20} \left(\frac {100}{t+1} + 5\right)\, \mathrm{d}t & = \left[100 \ln|t+1| +5t \right]_{10}^{20}\\ & =100 \ln|20+1| +5(20) -(100 \ln|10+1| +5(10) )\\ & =100 \ln(21) -100 \ln(11) +50\\ \implies \displaystyle \int_{10}^{20} \left(\frac {100}{t+1} + 5\right)\, \mathrm{d}t& \approx 114.663\\ \end{align*} {/eq}

We multiply them by {eq}1000 {/eq} since the rate is in thousands of barrels per year: {eq}74.315(1000) = 74315 {/eq} and {eq}114.663(1000) = 114663 {/eq}.

Thus, the field will produce approximately {eq}74315 {/eq} of barrels per year during the first year of production and {eq}114663 {/eq} barrels per year from the end of the 10th year to the end of the 20th year of production.