By: David Bradshaw

The velocity is constant from x=2 to x=8 because the slope does not change during this period. From x=2 to x=8, the velocity is 1.

A=lw A= (1) x (6) A= 6 square units A=D D= 6 miles

To find the distance a particle traveled, we find the area during the given time. This area is from x=2 to x=8. The area and the distance from x=2 to x=8 are the same.

Average velocity is (f(b)+ f(a))/2 From x=8 to x=10, the average velocity is (1-0)/2 = ½ miles per minute ½ miles per minute x 2 minutes = 1 mile traveled from x=8 to x=10.

A = ½ bh A = ½ (2)(1) = 1 mile traveled

The area from x=0 to x=10 most closely resembles a trapezoid.

(16/2) x 1 = 8 miles traveled

The actual area under the curve is 7 2/3 miles. My approximation in Q8 is too big.

Area of OUG =.19 miles Area of UADG =.96 miles

= 7.75 miles traveled (approximation) 7 2/3 miles traveled (area under curve) The approximation is too big.

My approximation approved significantly. It is within.1 miles of the real area, as opposed to my original approximation, which was 1/3 miles too big.