# 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.

*print*Print*list*Cite

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.**