Answer :
To approach this problem, let’s break it down step-by-step.
### Part (a): Using 4 Subintervals
Given the velocity function: [tex]\( v(t) = 2t^2 + 1 \)[/tex] for [tex]\( 0 \leq t \leq 4 \)[/tex].
1. Divide the interval [tex]\([0,4]\)[/tex] into 4 subintervals:
- [tex]\([0, 1]\)[/tex]
- [tex]\([1, 2]\)[/tex]
- [tex]\([2, 3]\)[/tex]
- [tex]\([3, 4]\)[/tex]
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 1]\)[/tex] is [tex]\( t = 0.5 \)[/tex]
- Midpoint of [tex]\([1, 2]\)[/tex] is [tex]\( t = 1.5 \)[/tex]
- Midpoint of [tex]\([2, 3]\)[/tex] is [tex]\( t = 2.5 \)[/tex]
- Midpoint of [tex]\([3, 4]\)[/tex] is [tex]\( t = 3.5 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity over each subinterval:
- [tex]\( v(0.5) = 2(0.5)^2 + 1 = 2(0.25) + 1 = 0.5 + 1 = 1.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.5) = 2(1.5)^2 + 1 = 2(2.25) + 1 = 4.5 + 1 = 5.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.5) = 2(2.5)^2 + 1 = 2(6.25) + 1 = 12.5 + 1 = 13.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.5) = 2(3.5)^2 + 1 = 2(12.25) + 1 = 24.5 + 1 = 25.5 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Displacement is the velocity times the time interval length. Each subinterval length is 1 second.
- [tex]\([0, 1]\)[/tex]: [tex]\( 1.5 \times 1 = 1.5 \, \text{ft} \)[/tex]
- [tex]\([1, 2]\)[/tex]: [tex]\( 5.5 \times 1 = 5.5 \, \text{ft} \)[/tex]
- [tex]\([2, 3]\)[/tex]: [tex]\( 13.5 \times 1 = 13.5 \, \text{ft} \)[/tex]
- [tex]\([3, 4]\)[/tex]: [tex]\( 25.5 \times 1 = 25.5 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 1.5 + 5.5 + 13.5 + 25.5 = 46 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 4 \)[/tex] subintervals is [tex]\( 46 \, \text{ft} \)[/tex].
### Part (b): Using 8 Subintervals
1. Divide the interval [tex]\([0, 4]\)[/tex] into 8 subintervals:
- Each subinterval is [tex]\([0, 0.5]\)[/tex], [tex]\([0.5, 1]\)[/tex], [tex]\([1, 1.5]\)[/tex], [tex]\([1.5, 2]\)[/tex], [tex]\([2, 2.5]\)[/tex], [tex]\([2.5, 3]\)[/tex], [tex]\([3, 3.5]\)[/tex], [tex]\([3.5, 4]\)[/tex].
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 0.5]\)[/tex] is [tex]\( t = 0.25 \)[/tex]
- Midpoint of [tex]\([0.5, 1]\)[/tex] is [tex]\( t = 0.75 \)[/tex]
- Midpoint of [tex]\([1, 1.5]\)[/tex] is [tex]\( t = 1.25 \)[/tex]
- Midpoint of [tex]\([1.5, 2]\)[/tex] is [tex]\( t = 1.75 \)[/tex]
- Midpoint of [tex]\([2, 2.5]\)[/tex] is [tex]\( t = 2.25 \)[/tex]
- Midpoint of [tex]\([2.5, 3]\)[/tex] is [tex]\( t = 2.75 \)[/tex]
- Midpoint of [tex]\([3, 3.5]\)[/tex] is [tex]\( t = 3.25 \)[/tex]
- Midpoint of [tex]\([3.5, 4]\)[/tex] is [tex]\( t = 3.75 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity:
- [tex]\( v(0.25) = 2(0.25)^2 + 1 = 2(0.0625) + 1 = 0.125 + 1 = 1.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(0.75) = 2(0.75)^2 + 1 = 2(0.5625) + 1 = 1.125 + 1 = 2.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.25) = 2(1.25)^2 + 1 = 2(1.5625) + 1 = 3.125 + 1 = 4.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.75) = 2(1.75)^2 + 1 = 2(3.0625) + 1 = 6.125 + 1 = 7.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.25) = 2(2.25)^2 + 1 = 2(5.0625) + 1 = 10.125 + 1 = 11.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.75) = 2(2.75)^2 + 1 = 2(7.5625) + 1 = 15.125 + 1 = 16.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.25) = 2(3.25)^2 + 1 = 2(10.5625) + 1 = 21.125 + 1 = 22.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.75) = 2(3.75)^2 + 1 = 2(14.0625) + 1 = 28.125 + 1 = 29.125 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Each subinterval length is 0.5 seconds.
- [tex]\([0, 0.5]\)[/tex]: [tex]\( 1.125 \times 0.5 = 0.5625 \, \text{ft} \)[/tex]
- [tex]\([0.5, 1]\)[/tex]: [tex]\( 2.125 \times 0.5 = 1.0625 \, \text{ft} \)[/tex]
- [tex]\([1, 1.5]\)[/tex]: [tex]\( 4.125 \times 0.5 = 2.0625 \, \text{ft} \)[/tex]
- [tex]\([1.5, 2]\)[/tex]: [tex]\( 7.125 \times 0.5 = 3.5625 \, \text{ft} \)[/tex]
- [tex]\([2, 2.5]\)[/tex]: [tex]\( 11.125 \times 0.5 = 5.5625 \, \text{ft} \)[/tex]
- [tex]\([2.5, 3]\)[/tex]: [tex]\( 16.125 \times 0.5 = 8.0625 \, \text{ft} \)[/tex]
- [tex]\([3, 3.5]\)[/tex]: [tex]\( 22.125 \times 0.5 = 11.0625 \, \text{ft} \)[/tex]
- [tex]\([3.5, 4]\)[/tex]: [tex]\( 29.125 \times 0.5 = 14.5625 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 0.5625 + 1.0625 + 2.0625 + 3.5625 + 5.5625 + 8.0625 + 11.0625 + 14.5625 = 46.5 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 8 \)[/tex] subintervals is [tex]\( 46.5 \, \text{ft} \)[/tex].
### Part (a): Using 4 Subintervals
Given the velocity function: [tex]\( v(t) = 2t^2 + 1 \)[/tex] for [tex]\( 0 \leq t \leq 4 \)[/tex].
1. Divide the interval [tex]\([0,4]\)[/tex] into 4 subintervals:
- [tex]\([0, 1]\)[/tex]
- [tex]\([1, 2]\)[/tex]
- [tex]\([2, 3]\)[/tex]
- [tex]\([3, 4]\)[/tex]
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 1]\)[/tex] is [tex]\( t = 0.5 \)[/tex]
- Midpoint of [tex]\([1, 2]\)[/tex] is [tex]\( t = 1.5 \)[/tex]
- Midpoint of [tex]\([2, 3]\)[/tex] is [tex]\( t = 2.5 \)[/tex]
- Midpoint of [tex]\([3, 4]\)[/tex] is [tex]\( t = 3.5 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity over each subinterval:
- [tex]\( v(0.5) = 2(0.5)^2 + 1 = 2(0.25) + 1 = 0.5 + 1 = 1.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.5) = 2(1.5)^2 + 1 = 2(2.25) + 1 = 4.5 + 1 = 5.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.5) = 2(2.5)^2 + 1 = 2(6.25) + 1 = 12.5 + 1 = 13.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.5) = 2(3.5)^2 + 1 = 2(12.25) + 1 = 24.5 + 1 = 25.5 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Displacement is the velocity times the time interval length. Each subinterval length is 1 second.
- [tex]\([0, 1]\)[/tex]: [tex]\( 1.5 \times 1 = 1.5 \, \text{ft} \)[/tex]
- [tex]\([1, 2]\)[/tex]: [tex]\( 5.5 \times 1 = 5.5 \, \text{ft} \)[/tex]
- [tex]\([2, 3]\)[/tex]: [tex]\( 13.5 \times 1 = 13.5 \, \text{ft} \)[/tex]
- [tex]\([3, 4]\)[/tex]: [tex]\( 25.5 \times 1 = 25.5 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 1.5 + 5.5 + 13.5 + 25.5 = 46 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 4 \)[/tex] subintervals is [tex]\( 46 \, \text{ft} \)[/tex].
### Part (b): Using 8 Subintervals
1. Divide the interval [tex]\([0, 4]\)[/tex] into 8 subintervals:
- Each subinterval is [tex]\([0, 0.5]\)[/tex], [tex]\([0.5, 1]\)[/tex], [tex]\([1, 1.5]\)[/tex], [tex]\([1.5, 2]\)[/tex], [tex]\([2, 2.5]\)[/tex], [tex]\([2.5, 3]\)[/tex], [tex]\([3, 3.5]\)[/tex], [tex]\([3.5, 4]\)[/tex].
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 0.5]\)[/tex] is [tex]\( t = 0.25 \)[/tex]
- Midpoint of [tex]\([0.5, 1]\)[/tex] is [tex]\( t = 0.75 \)[/tex]
- Midpoint of [tex]\([1, 1.5]\)[/tex] is [tex]\( t = 1.25 \)[/tex]
- Midpoint of [tex]\([1.5, 2]\)[/tex] is [tex]\( t = 1.75 \)[/tex]
- Midpoint of [tex]\([2, 2.5]\)[/tex] is [tex]\( t = 2.25 \)[/tex]
- Midpoint of [tex]\([2.5, 3]\)[/tex] is [tex]\( t = 2.75 \)[/tex]
- Midpoint of [tex]\([3, 3.5]\)[/tex] is [tex]\( t = 3.25 \)[/tex]
- Midpoint of [tex]\([3.5, 4]\)[/tex] is [tex]\( t = 3.75 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity:
- [tex]\( v(0.25) = 2(0.25)^2 + 1 = 2(0.0625) + 1 = 0.125 + 1 = 1.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(0.75) = 2(0.75)^2 + 1 = 2(0.5625) + 1 = 1.125 + 1 = 2.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.25) = 2(1.25)^2 + 1 = 2(1.5625) + 1 = 3.125 + 1 = 4.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.75) = 2(1.75)^2 + 1 = 2(3.0625) + 1 = 6.125 + 1 = 7.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.25) = 2(2.25)^2 + 1 = 2(5.0625) + 1 = 10.125 + 1 = 11.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.75) = 2(2.75)^2 + 1 = 2(7.5625) + 1 = 15.125 + 1 = 16.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.25) = 2(3.25)^2 + 1 = 2(10.5625) + 1 = 21.125 + 1 = 22.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.75) = 2(3.75)^2 + 1 = 2(14.0625) + 1 = 28.125 + 1 = 29.125 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Each subinterval length is 0.5 seconds.
- [tex]\([0, 0.5]\)[/tex]: [tex]\( 1.125 \times 0.5 = 0.5625 \, \text{ft} \)[/tex]
- [tex]\([0.5, 1]\)[/tex]: [tex]\( 2.125 \times 0.5 = 1.0625 \, \text{ft} \)[/tex]
- [tex]\([1, 1.5]\)[/tex]: [tex]\( 4.125 \times 0.5 = 2.0625 \, \text{ft} \)[/tex]
- [tex]\([1.5, 2]\)[/tex]: [tex]\( 7.125 \times 0.5 = 3.5625 \, \text{ft} \)[/tex]
- [tex]\([2, 2.5]\)[/tex]: [tex]\( 11.125 \times 0.5 = 5.5625 \, \text{ft} \)[/tex]
- [tex]\([2.5, 3]\)[/tex]: [tex]\( 16.125 \times 0.5 = 8.0625 \, \text{ft} \)[/tex]
- [tex]\([3, 3.5]\)[/tex]: [tex]\( 22.125 \times 0.5 = 11.0625 \, \text{ft} \)[/tex]
- [tex]\([3.5, 4]\)[/tex]: [tex]\( 29.125 \times 0.5 = 14.5625 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 0.5625 + 1.0625 + 2.0625 + 3.5625 + 5.5625 + 8.0625 + 11.0625 + 14.5625 = 46.5 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 8 \)[/tex] subintervals is [tex]\( 46.5 \, \text{ft} \)[/tex].
