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# 数学代写|微积分代写Calculus代考|Displacement Versus Distance Traveled

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## 数学代写|微积分代写Calculus代考|Displacement Versus Distance Traveled

If an object with position function $s(t)$ moves along a coordinate line without changing direction, we can calculate the total distance it travels from $t=a$ to $t=b$ by summing the distance traveled over small intervals, as in Example 1. If the object reverses direction one or more times during the trip, then we need to use the object’s speed $|v(t)|$, which is the absolute value of its velocity function, $v(t)$, to find the total distance traveled. Using the velocity itself, as in Example 1, gives instead an estimate to the object’s displacement, $s(b)-s(a)$, the difference between its initial and final positions. To see the difference, think about what happens when you walk a mile from your home and then walk back. The total distance traveled is two miles, but your displacement is zero, because you end up back where you started.

To see why using the velocity function in the summation process gives an estimate to the displacement, partition the time interval $[a, b]$ into small enough equal subintervals $\Delta t$ so that the object’s velocity does not change very much from time $t_{k-1}$ to $t_k$. Then $v\left(t_k\right)$ gives a good approximation of the velocity throughout the interval. Accordingly, the change in the object’s position coordinate, which is its displacement during the time interval, is about
$$v\left(t_k\right) \Delta t$$
The change is positive if $v\left(t_k\right)$ is positive and negative if $v\left(t_k\right)$ is negative.
In either case, the distance traveled by the object during the subinterval is about
$$\left|v\left(t_k\right)\right| \Delta t$$
The total distance traveled over the time interval is approximately the sum
$$\left|v\left(t_1\right)\right| \Delta t+\left|v\left(t_2\right)\right| \Delta t+\cdots+\left|v\left(t_n\right)\right| \Delta t$$
We will revisit these ideas in Section 5.4.

## 数学代写|微积分代写Calculus代考|Average Value of a Nonnegative Continuous Function

The average value of a collection of $n$ numbers $x_1, x_2, \ldots, x_n$ is obtained by adding them together and dividing by $n$. But what is the average value of a continuous function $f$ on an interval $[a, b]$ ? Such a function can assume infinitely many values. For example, the temperature at a certain location in a town is a continuous function that goes up and down each day. What does it mean to say that the average temperature in the town over the course of a day is 73 degrees?

When a function is constant, this question is easy to answer. A function with constant value $c$ on an interval $[a, b]$ has average value $c$. When $c$ is positive, its graph over $[a, b]$ gives a rectangle of height $c$. The average value of the function can then be interpreted geometrically as the area of this rectangle divided by its width $b-a$ (see Figure 5.6a).

What if we want to find the average value of a nonconstant function, such as the function $g$ in Figure 5.6b? We can think of this graph as a snapshot of the height of some water that is sloshing around in a tank between enclosing walls at $x=a$ and $x=b$. As the water moves, its height over each point changes, but its average height remains the same. To get the average height of the water, we let it settle down until it is level and its height is constant. The resulting height $c$ equals the area under the graph of $g$ divided by $b-a$. We are led to define the average value of a nonnegative function on an interval $[a, b]$ to be the area under its graph divided by $b-a$. For this definition to be valid, we need a precise understanding of what is meant by the area under a graph. This will be obtained in Section 5.3 , but for now we look at an example.

## 数学代写|微积分代写Calculus代考|Displacement Versus Distance Traveled

$$v\left(t_k\right) \Delta t$$

$$\left|v\left(t_k\right)\right| \Delta t$$

$$\left|v\left(t_1\right)\right| \Delta t+\left|v\left(t_2\right)\right| \Delta t+\cdots+\left|v\left(t_n\right)\right| \Delta t$$

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