MHB Need help with velocity problem that I have no idea how to do

[sma14]

The velocity of an object is given by the following function defined on a specific interval. Approximate the displacement of the object on the interval by subdiving the interval into the indicated number of subintervals. Use the left endpoint of each subinterval to compute the height of the recentangles
v=1(3t+2) (m/s) for 0 ≤t ≤8, n=4

 

[topsquark]

The velocity of an object is given by the following function defined on a specific interval. Approximate the displacement of the object on the interval by subdiving the interval into the indicated number of subintervals. Use the left endpoint of each subinterval to compute the height of the recentangles
v=1(3t+2) (m/s) for 0 ≤t ≤8, n=4

What have you been able to do so far? This is most easily discussed as a graphing problem.

-Dan

 

[HOI]

That "1" in front of the parentheses is unnecessary and looks strange. Are you sure it isn't v= 1/(3t+ 2)?

In any case, you are told exactly what to do in the problem:

Approximate the displacement of the object on the interval by subdiving the interval into the indicated number of subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles
v=1(3t+2) (m/s) for 0 ≤t ≤8, n=4

Dividing 0 to 8 into 4 subintervals means [0, 2], [2, 4], [4, 6], and [6, 8]. Although the height of the function varies continuously, you are told to "use the left endpoint of each subinterval to compte the height". So you are approximating the area under the curve by 4 rectangles, each with length 2 and heights calculated at t= 0, 2, 4, and 6.

If v really is 1(3t+ 2) then those heights are 1(3(0)+ 2)= 2, 1(3(2)+ 2)= 8, 1(3(4)+ 2)= 14, and 1(3(6)+ 2)= 20. What are the areas of those four rectangles? What is the total area. But if v= 1/(t+ 2), the heights are 1/(0+ 2), 1/(2+ 2), 1/(4+ 2), and 1/(6+ 2). What are the areas of those rectangles? What is the total area?