for y=√x the midpoints with 4 rectangles are 1/8, 3/8, 5/8, 7/8
So, the area is 1/4 (√(1/8) + √(3/8) + √(5/8) + √(7/8)) = (1+√3+√5+√7)/(8√2)
I think you mean
v(t)=200 - 32t
not sure why you'd want an estimate for velocity, since you have the formula. Clearly, 200 is an upper estimate, since v decreases with time.
For distance, using right-side rectangles of width 1, the distance would be greater than
1(v(1)+v(2)+v(3)+v(4)+v(5))
1. Using midpoint method, estimate the area under the graph of y= sqrt x between x = 0 to x = 1, using 4 rectangles.
3. A soccer ball is kicked straight up into the air from the ground with an initial velocity of 200 ft/s. The velocity of the ball at any moment is shown with v(t)=200 - 32 ft/s.
How would you find:
a) an upper estimate for the velocity after 5 sec have elapsed.
b) a lower estimate for the height attained after 5 sec.
1 answer