In Lesson 4.2.3 you should have noticed a close relationship between derivatives and integrals just like with velocity and distance. In particular, you have seen that if you know the velocity, $v\left(t\right)$, at time $t$ then you can compute the distance traveled from $t = 0$ to $t = x$ by using $s ( x ) = \int _ { 0 } ^ { x } v ( t ) d t$.

1. If $v\left(t\right) = 2t + 5$ in miles per hour, how far has the car traveled after $2$ hours? After $5$ hours? After $x$ hours?

   Hint (a):

   For the first 2 hours: $\int_{0}^{2}{(2t+5)dt}=$

   More Help (a):

   See the homework help for problem 4-99.

2. How far did the car travel between $2$ and $4$ hours?

   Hint (b):

   $s\left(4\right) – s\left(2\right) =$

3. In Chapter 3 you considered the instantaneous rate of change (derivative) of any function. Explain why you expect the derivative, $s^\prime\left(t\right)$, of a distance function of a car to be the velocity function, $v\left(t\right)$, of the car.

   Hint (c):

   Since s is a distance function, what does the slope of a distance function represent?