Math / Calculus

QuestionThe marginal cost of oil production, in dollars per barrel, is represented by $\mathrm{C}^{\prime}(\mathrm{x})$ , where x is the number of barrels of oil produced. Report the units of $\int_{600} \mathrm{C}^{\prime}(\mathrm{x}) \mathrm{dx}$ and interpret what the integral means.
The units of $\int_{600}^{640} C^{\prime}(x) d x$ are $\square$

Studdy Solution

STEP 1

1. The function $C'(x)$ represents the marginal cost of producing oil in dollars per barrel.
2. The variable $x$ represents the number of barrels of oil produced.
3. The integral $\int_{600}^{640} C^{\prime}(x) \, dx$ involves calculating the total change in cost over a specific range of production.

STEP 2

1. Determine the units of the integral.
2. Interpret the meaning of the integral.

STEP 3

To determine the units of $\int_{600}^{640} C^{\prime}(x) \, dx$ , we need to consider the units of the integrand $C'(x)$ and the differential $dx$ .
- The units of $C'(x)$ are dollars per barrel. - The units of $dx$ are barrels.
When you multiply these together in the context of an integral, the units of the integral become:
$\text{dollars per barrel} \times \text{barrels} = \text{dollars}$
Therefore, the units of $\int_{600}^{640} C^{\prime}(x) \, dx$ are dollars.

STEP 4

Interpret the meaning of the integral $\int_{600}^{640} C^{\prime}(x) \, dx$ .
The integral of a marginal cost function over a specific interval gives the total change in cost associated with producing the additional barrels of oil within that interval. Specifically, $\int_{600}^{640} C^{\prime}(x) \, dx$ represents the total increase in cost when production increases from 600 barrels to 640 barrels.
The units of $\int_{600}^{640} C^{\prime}(x) \, dx$ are $\boxed{\text{dollars}}$ .

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