A car starts moving at time t=0 and goes faster and faster. Its velocity is shown in the following table.

Question

t (seconds)	0	3	6	9	12
Velocity (ft/sec)	0	8	20	39	70

A. Estimate how far the car traveled during the first 12 seconds using the left-hand sums with 4 subdivisions. Answer: ____ft

B. Now estimate how far the car traveled during the first 12 seconds using the right-hand sums with four subdivisions. Answer: ____ft

Determine which of the two is underestimate: (choose A or B)

Answer 1

(A) S ≈ L₄ = (0 + 8 + 20 + 39)·3 = 201 ft

(B) S ≈ R₄ = (8 + 20 + 39 + 70)·3 = 411 ft

Answer 2

A. For the Reimann Left-Hand Sum, the formula will be:

∫^b_a f(x)dx ≈ Δx(f(x₀)+f(x₁)+f(x₂)+...+f(x_n-1))

Where: Δx = (b-a)/n

Given:

b=12, a=0, n=4

Therefore:

Δt = (12-0)/4 = 3

The estimate for the distance traveled by the car is:

∫¹²₀v(t)dt ≈ Δt(v(t₀)+v(t₁)+v(t₂)+v(t₃))

≈ 3(v(0)+v(3)+v(6)+v(9))

≈ 3(0+8+20+39)

≈ 211 ft.

B. For the Reimann Right-Hand Sum, the formula will be:

∫^b_a f(x)dx ≈ Δx(f(x₁)+f(x₂)+f(x₃)+...+f(x_n))

Where: Δx = (b-a)/n

The estimate for the distance traveled by the car is:

∫¹²₀v(t)dt ≈ Δt(v(t₁)+v(t₂)+v(t₃)+v(t₄))

≈ 3(v(3)+v(6)+v(9)+v(12))

≈ 3(8+20+39+70)

≈ 411 ft.

Can you now determine the underestimate? I think it's very obvious.