# Using integrals solve the following question. Any assistance would be greatly appreciated.A robot leaving a spaceship has velocity given by v(t)= −0.4t^2 + 3t, where v(t) is inkilometers/hour...

Using integrals solve the following question. Any assistance would be greatly appreciated.

A robot leaving a spaceship has velocity given by v(t)= −0.4t^2
+ 3t, where v(t) is inkilometers/hour and t is the number of hours since it left the space ship.

a. Estimate the distance travelled during the first 5 hours using 4 subintervals of equal length and the midpoint of each subinterval to construct rectangles.

b. Find the distance travelled exactly.

embizze | High School Teacher | (Level 1) Educator Emeritus

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Given `v(t)=-0.4t^2+3t` :

(1) An approximation for the distance travelled in the first five hours can be found by approximating the area under the curve from 0 to 5.

Dividing the interval into four equal subintervals gives the endpoints of the subintervals as 0,5/4,5/2,15/4,5.

The midpoints of the four subintervals are `5/8,15/8,25/8,35/8` . The width of each subinterval is 5/4.

An approximation for the area under the curve using these subintervals is:

`sum_(i=1)^(5) f(c_i)Deltax_i` where `Delta x_i=5/4` and `c_i` is the midpoint of the `i^(th)` subinterval -- factoring out the 5/4 we get

`5/4(f(5/8)+f(15/8)+f(25/8)+f(35/8))`

`~~5/4(1.7188+4.2188+5.4688+5.4688)`

`~~5/4(16.8752)~~21.094`

Thus the approximation for the area under the curve is 21.094 and the approximate distance travelled is 21.094km.

(2) The exact distance is the area under the curve. This is found using a Riemann sum or integrating the function:

`A=int_0^5(-.04t^2+3t)dt`

`=(-.4t^3)/3+(3t^2)/2 |_0^5`

`=20.8bar(3)=125/6`

So the exact distance travelled is 125/6km.