Relation between velocity and area

Suppose you're reading a car magazine and there is an article about a new sports car that has this table in it:

Time (seconds)	0	1	2	3	4	5	6
Speed (mph)	0	5	15	25	40	50	60

They claim the car drove $1/8$th of a mile after $6$ seconds, but this just ``feels'' wrong... Hmmm... Let's estimate the distance driven using the formula

    distance $$=$$    rate $$\times$$    time$$.$$

We overestimate by assuming the velocity is a constant equal to the max on each interval:

   estimate $$=5 \cdot 1 + 15 \cdot 1 + 25 \cdot 1 + 40 \cdot 1 + 50 \cdot 1 + 60 \cdot 1 = \frac{195}{3600}$$    miles $$= 0.054...$$

(Note: there are $3600$ seconds in an hour.) But $1/8 \sim 0.125$, so the article is inconsistent. (Doesn't this sort of thing just bug you? By learning calculus you'll be able to double-check things like this much more easily.)

Insight! The formula for the estimate of distance traveled above looks exactly like an approximation for the area under the graph of the speed of the car! In fact, if an object has velocity $v(t)$ at time $t$, then the net change in position from time $a$ to $b$ is

$$\int_{a}^b v(t) dt.$$

We'll come back to this observation frequently.