Question

a company manufactures mountain bikes. the research department produced the marginal cost function $c(x)=700-\frac{x}{3},0leq xleq900$, where $c(x)$ is in dollars and $x$ is the number of bikes produced per month. compute the increase in cost going from a production level of 300 bikes per month to 900 bikes per month. set up a definite integral and evaluate it.

Explanation:

Step1: Set up the definite integral

The increase in cost is given by the definite integral of the marginal cost function \(C^{\prime}(x)\) from \(x = 300\) to \(x=900\). So, we set up the integral \(\int_{300}^{900}(700-\frac{x}{3})dx\).

Step2: Integrate term - by - term

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and \(\int a dx=ax + C\) (where \(a\) is a constant):
\(\int(700-\frac{x}{3})dx=\int700dx-\frac{1}{3}\int xdx=700x-\frac{1}{3}\times\frac{x^{2}}{2}+C=700x-\frac{x^{2}}{6}+C\).

Step3: Evaluate the definite integral

By the fundamental theorem of calculus \(\int_{a}^{b}F^{\prime}(x)dx=F(b)-F(a)\), where \(F(x)=700x-\frac{x^{2}}{6}\).
\(F(900)-F(300)=(700\times900-\frac{900^{2}}{6})-(700\times300-\frac{300^{2}}{6})\).
First, calculate \(700\times900 = 630000\), \(\frac{900^{2}}{6}=\frac{810000}{6}=135000\), \(700\times300 = 210000\), \(\frac{300^{2}}{6}=\frac{90000}{6}=15000\).
Then \(F(900)-F(300)=(630000 - 135000)-(210000 - 15000)\).
\(=495000-195000\).

Answer:

\(300000\) dollars.