Question

exercises 1.3 the fundamental theorem of calculus
score: 245/260 answered: 25/26
question 26
the traffic flow rate (cars per hour) across an intersection is $r(t)=200 + 1000t - 150t^2$, where $t$ is in hours, and $t = 0$ is 6am. how many cars pass through the intersection between 6 am and 8 am?
cars
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Explanation:

Step1: Determine the time interval

From 6 am (\(t = 0\)) to 8 am, the time elapsed is \(t=2\) hours. We need to find the integral of the rate function \(r(t)=200 + 1000t-150t^{2}\) from \(t = 0\) to \(t = 2\). The formula for the total number of cars \(N\) is \(N=\int_{a}^{b}r(t)dt\), where \(a = 0\) and \(b = 2\).

Step2: Integrate the rate function

First, find the antiderivative of \(r(t)\). The antiderivative of \(200\) with respect to \(t\) is \(200t\), the antiderivative of \(1000t\) is \(\frac{1000t^{2}}{2}=500t^{2}\), and the antiderivative of \(- 150t^{2}\) is \(\frac{-150t^{3}}{3}=- 50t^{3}\). So the antiderivative \(F(t)=200t + 500t^{2}-50t^{3}\).

Step3: Evaluate the definite integral

Using the Fundamental Theorem of Calculus, \(\int_{0}^{2}r(t)dt=F(2)-F(0)\).

First, calculate \(F(2)\):
\(F(2)=200(2)+500(2)^{2}-50(2)^{3}=400 + 500\times4-50\times8=400 + 2000 - 400\)

Then, calculate \(F(0)\):
\(F(0)=200(0)+500(0)^{2}-50(0)^{3}=0\)

So \(\int_{0}^{2}r(t)dt=(400 + 2000 - 400)-0=2000\)

Answer:

2000