Can you explain how to do these three problems?

Step-by-step explanation:

a) The distance is the integral of the velocity vs. time graph.  We can approximate the distance using a left hand Riemann sum.  That means for each interval, use the velocity at the beginning of the interval.  Don't forget to convert mi/hr to mi/s.

d₁ = (10 s − 0 s) (183.9 mi/hr × 1 hr / 3600 s) = 0.5108 mi

d₂ = (20 s − 10 s) (169.0 mi/hr × 1 hr / 3600 s) = 0.4694 mi

d₃ = (30 s − 20 s) (105.6 mi/hr × 1 hr / 3600 s) = 0.2933 mi

d₄ = (40 s − 30 s) (99.8 mi/hr × 1 hr / 3600 s) = 0.2772 mi

d₅ = (50 s − 40 s) (124.5 mi/hr × 1 hr / 3600 s) = 0.3458 mi

d₆ = (60 s − 50 s) (177.1 mi/hr × 1 hr / 3600 s) = 0.4936 mi

d = 0.5108 + 0.4694 + 0.2933 + 0.2772 + 0.3458 + 0.4936

d = 2.390 miles

b) Do the same as part a, but this time, use a right hand Riemann sum.  Instead of using the velocity at the beginning of the interval, use the velocity at the end of the interval.

d₁ = (10 s − 0 s) (169.0 mi/hr × 1 hr / 3600 s) = 0.4694 mi

d₂ = (20 s − 10 s) (105.6 mi/hr × 1 hr / 3600 s) = 0.2933 mi

d₃ = (30 s − 20 s) (99.8 mi/hr × 1 hr / 3600 s) = 0.2772 mi

d₄ = (40 s − 30 s) (124.5 mi/hr × 1 hr / 3600 s) = 0.3458 mi

d₅ = (50 s − 40 s) (177.1 mi/hr × 1 hr / 3600 s) = 0.4936 mi

d₆ = (60 s − 50 s) (175.6 mi/hr × 1 hr / 3600 s) = 0.4878 mi

d = 0.4694 + 0.2933 + 0.2772 + 0.3458 + 0.4936 + 0.4878

d = 2.367 miles

c) The velocity decreases from 0 s to 30 s, but then increases from 30 s to 50 s, and then decreases from 50 s to 60 s.

Since the velocity doesn't consistently increase or decrease, these Riemann sums are neither lower nor upper sums.