Answer :
To determine the number of cars that pass through the intersection between 6 am and 7 am, we need to integrate the traffic flow rate function [tex]\( r(t) = 500 + 1000t - 210t^2 \)[/tex] over the interval from [tex]\( t = 0 \)[/tex] to [tex]\( t = 1 \)[/tex].
1. Set up the integral:
[tex]\[ \int_{0}^{1} (500 + 1000t - 210t^2) \, dt \][/tex]
2. Integrate the function:
We need to integrate each term individually.
- The integral of [tex]\( 500 \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \int 500 \, dt = 500t \][/tex]
- The integral of [tex]\( 1000t \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \int 1000t \, dt = 1000 \int t \, dt = 1000 \left( \frac{t^2}{2} \right) = 500t^2 \][/tex]
- The integral of [tex]\( -210t^2 \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \int -210t^2 \, dt = -210 \int t^2 \, dt = -210 \left( \frac{t^3}{3} \right) = -70t^3 \][/tex]
3. Combine the results:
Using the results from the individual integrals, we get:
[tex]\[ \int_{0}^{1} (500 + 1000t - 210t^2) \, dt = \left[ 500t + 500t^2 - 70t^3 \right]_{0}^{1} \][/tex]
4. Evaluate the definite integral:
Now, we evaluate the expression from [tex]\( t = 0 \)[/tex] to [tex]\( t = 1 \)[/tex]:
- At [tex]\( t = 1 \)[/tex]:
[tex]\[ 500(1) + 500(1)^2 - 70(1)^3 = 500 + 500 - 70 = 930 \][/tex]
- At [tex]\( t = 0 \)[/tex]:
[tex]\[ 500(0) + 500(0)^2 - 70(0)^3 = 0 + 0 - 0 = 0 \][/tex]
5. Subtract the lower bound result from the upper bound result:
[tex]\[ 930 - 0 = 930 \][/tex]
Therefore, the number of cars that pass through the intersection between 6 am and 7 am is [tex]\( \boxed{930} \)[/tex].
1. Set up the integral:
[tex]\[ \int_{0}^{1} (500 + 1000t - 210t^2) \, dt \][/tex]
2. Integrate the function:
We need to integrate each term individually.
- The integral of [tex]\( 500 \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \int 500 \, dt = 500t \][/tex]
- The integral of [tex]\( 1000t \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \int 1000t \, dt = 1000 \int t \, dt = 1000 \left( \frac{t^2}{2} \right) = 500t^2 \][/tex]
- The integral of [tex]\( -210t^2 \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \int -210t^2 \, dt = -210 \int t^2 \, dt = -210 \left( \frac{t^3}{3} \right) = -70t^3 \][/tex]
3. Combine the results:
Using the results from the individual integrals, we get:
[tex]\[ \int_{0}^{1} (500 + 1000t - 210t^2) \, dt = \left[ 500t + 500t^2 - 70t^3 \right]_{0}^{1} \][/tex]
4. Evaluate the definite integral:
Now, we evaluate the expression from [tex]\( t = 0 \)[/tex] to [tex]\( t = 1 \)[/tex]:
- At [tex]\( t = 1 \)[/tex]:
[tex]\[ 500(1) + 500(1)^2 - 70(1)^3 = 500 + 500 - 70 = 930 \][/tex]
- At [tex]\( t = 0 \)[/tex]:
[tex]\[ 500(0) + 500(0)^2 - 70(0)^3 = 0 + 0 - 0 = 0 \][/tex]
5. Subtract the lower bound result from the upper bound result:
[tex]\[ 930 - 0 = 930 \][/tex]
Therefore, the number of cars that pass through the intersection between 6 am and 7 am is [tex]\( \boxed{930} \)[/tex].
