Question

a company manufactures mountain bikes. the research department produced the marginal cost function ( c^{prime}(x)=600-\frac{x}{3}, 0 leq x leq 900 ), where ( c^{prime}(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 0 bikes per month to 450 bikes per month. set up a definite integral and evaluate it. set up the definite integral.
( int_{0}^{square}(square) d x )

Explanation:

Step1: Recall the formula for cost increase

The increase in cost from \(x = a\) to \(x = b\) is given by \(\int_{a}^{b}C^{\prime}(x)dx\). Here, \(a = 0\), \(b = 450\), and \(C^{\prime}(x)=600-\frac{x}{3}\).
So the definite - integral is \(\int_{0}^{450}(600 - \frac{x}{3})dx\).

Step2: Integrate the function

We know that \(\int(600-\frac{x}{3})dx=\int600dx-\int\frac{x}{3}dx\).
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and \(\int kdx=kx + C\) (\(k\) is a constant):
\(\int600dx=600x\) and \(\int\frac{x}{3}dx=\frac{1}{3}\times\frac{x^{2}}{2}=\frac{x^{2}}{6}\).
So \(\int(600-\frac{x}{3})dx=600x-\frac{x^{2}}{6}+C\).

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus \(\int_{a}^{b}f(x)dx=F(b)-F(a)\), where \(F(x)\) is an antiderivative of \(f(x)\).
\(F(x)=600x-\frac{x^{2}}{6}\), then \(F(450)-F(0)=(600\times450-\frac{450^{2}}{6})-(600\times0-\frac{0^{2}}{6})\).
First, calculate \(600\times450 = 270000\).
Second, calculate \(\frac{450^{2}}{6}=\frac{202500}{6}=33750\).
Then \(F(450)-F(0)=270000 - 33750=236250\).

Answer:

The definite integral is \(\int_{0}^{450}(600-\frac{x}{3})dx\) and its value is \(236250\) dollars.