1. By: David Bradshaw

2. 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.

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

4. 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.

5. 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.

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

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

8. (16/2) x 1 = 8 miles traveled

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

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

11. .125 +.625 + 6 + 1 = 7.75 miles traveled (approximation) 7 2/3 miles traveled (area under curve) The approximation is too big.

12. 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.