An oil field is estimated to produce oil at a rate of R(t)...

An oil field is estimated to produce oil at a rate of $R(t)$ thousand barrels per month $t$ months from now, as given by
$$R(t)=10 t e^{-0.1 t}$$ Use an appropriate definite integral to find the total production (to the nearest thousand barrels) in the first year of operation.

Key Concepts

Definite Integral

A definite integral calculates the accumulation of a quantity over a continuous interval. In applications like this, it is used to totalize a rate function over time, providing the complete amount produced, consumed, or accumulated within that period.

Rate Function

A rate function represents the speed at which a quantity changes with respect to time. In the oil production example, the rate function indicates the number of barrels produced per month and integrating it over a specific time interval gives the total production.

Exponential Function

Exponential functions describe processes that have constant proportional growth or decay. When included in a rate function, such as an exponential decay factor, they model situations where increases or decreases occur continuously at a rate proportional to the current amount.

Limits of Integration

The limits of integration define the interval over which the accumulation is measured. In the context of the problem, setting the limits from the start to the end of the first year (0 to 12 months) ensures that the integration captures the total production during that period.

Transcript

00:01 R of t, which is the rate that oil is produced per month in thousands of barrels.

00:10 So we're asked to use a definite integral to find the total production in the first year of operation.

00:18 So if we're doing the first year of operation and t is the number of months, then our integral is going to be from 0 to 12.

00:31 So we're taking the integral from 0 to 12 of this function r of t.

00:36 So that's 10, t, e, negative 0 .1t.

00:44 Okay, so to do this, we're going to need to use integration by parts.

00:48 So we're going to let u be equal to t, which means our dv is b to the negative 0 .1 t, d t.

01:00 So du is going to be d t, and v is, is going to be e to the negative 0 .1t, but we need to have a negative 10 in a front to cancel out that negative 0 .1.

01:16 So now we can say this is equal to, and we'll just pull that 10 to the front because it's just a coefficient.

01:23 So that's 10, and then putting our integration by parts formula in, that's going to give us uv.

01:30 So that would be negative 10, t, e to the negative 0 .1 t, minus the integral of vdu.

01:41 So let's actually do 012 here first.

01:44 So that will be another negative, so that turns into a plus, and then now the integral from 0 to 12 of vdu, so that would be 10, t, e, sorry, vd, so there's no t, just 10b to negative 0 .1t, okay, so this is equal to 10, times, okay, let's substitute in our 12.

02:13 So this is going to be negative 120e to the negative 1 .2...

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