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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.1.57b
Capitolo 6, Problema 6.1.57b

55–58. Marginal cost Consider the following marginal cost functions.


b. Find the additional cost incurred in dollars when production is increased from 500 units to 550 units.


C′(x) = 300+10x−0.01x²

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Understand that the marginal cost function \(C\prime(x)\) represents the rate of change of the total cost with respect to the number of units produced, \(x\). To find the additional cost incurred when production increases from 500 to 550 units, we need to find the change in total cost over this interval.
Recall that the additional cost when increasing production from \(x = a\) to \(x = b\) can be found by integrating the marginal cost function over the interval \([a, b]\). In this case, we want to compute \(\int_{500}^{550} C\prime(x) \, dx\).
Set up the definite integral using the given marginal cost function: \(\int_{500}^{550} \left(300 + 10x - 0.01x^{2}\right) \, dx\).
Integrate the function term-by-term: - The integral of \(300\) with respect to \(x\) is \$300x$. - The integral of \$10x$ is \$5x^{2}$. - The integral of $-0.01x^{2}$ is \(-0.01 \times \frac{x^{3}}{3} = -\frac{0.01}{3} x^{3}\).
Evaluate the resulting antiderivative at the upper limit \(x=550\) and the lower limit \(x=500\), then subtract to find the total additional cost: \(\left[300x + 5x^{2} - \frac{0.01}{3} x^{3}\right]_{500}^{550} = \left(300 \times 550 + 5 \times 550^{2} - \frac{0.01}{3} \times 550^{3}\right) - \left(300 \times 500 + 5 \times 500^{2} - \frac{0.01}{3} \times 500^{3}\right)\).

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Marginal Cost Function

The marginal cost function, C′(x), represents the rate of change of the total cost with respect to the number of units produced. It indicates the additional cost of producing one more unit at production level x. Understanding this helps in estimating cost changes for small increases in production.
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Properties of Functions

Definite Integral for Accumulated Change

To find the total additional cost when production increases over an interval, integrate the marginal cost function over that range. The definite integral of C′(x) from x = 500 to x = 550 gives the exact increase in total cost for producing those extra 50 units.
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Definition of the Definite Integral

Evaluating Polynomial Integrals

Since the marginal cost function is a polynomial, integrating it involves applying power rules to each term. Accurate integration and evaluation at the bounds are essential to compute the total additional cost correctly, ensuring precise results for the production increase.
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Taylor Polynomials
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