t 0 4 8 12 16 20
vt 43 42 40 35 25 5
The following table gives the velocity v(t) (in feet per sec) at different instances of time t (in sec) of a particle moving along a horizontal axis.
t 0 4 8 12 16 20
v(t) 43 42 40 35 25 5
Estimate the distance traveled by the particle between t = 0 sec and t = 20 sec using 5 time intervals of equal length with
(a) left endpoints
Estimated distance = _____feet
(b) right endpoints
Estimated distance = _____feet
4 answers
left endpoints: 4*43 + 4*42 + ... + 4*25
right ends: 4*42 + 4*40 + ... + 4*5
right ends: 4*42 + 4*40 + ... + 4*5
hey thanks but i still don't get how you arrived to this answer?
consult your text, or do a web search. To approximate the area under a curve, draw rectangles of (in this case) width 4.
The vertical sides intersect the curve at various points. To approximate using left endpoints, make the height of the rectangle spanning x0-x1 the value of f(x0) - the left height.
Thus, the area of rectangle from xi to xi+1 is w*f(xi)
Using the right endpoints of the intervals, the area is w*f(xi+1)
Takes a lot of words, but any illustration should make it clear.
Why are we approximating the area under the curve? Because we are given the speed at various times. speed is the derivative of the position. The position is thus the integral of the speed over an interval.
The vertical sides intersect the curve at various points. To approximate using left endpoints, make the height of the rectangle spanning x0-x1 the value of f(x0) - the left height.
Thus, the area of rectangle from xi to xi+1 is w*f(xi)
Using the right endpoints of the intervals, the area is w*f(xi+1)
Takes a lot of words, but any illustration should make it clear.
Why are we approximating the area under the curve? Because we are given the speed at various times. speed is the derivative of the position. The position is thus the integral of the speed over an interval.
1 answer