Q. 7 - Calculus

Question

Intuitively, the average value of a velocity function should be the same as the average rate of change of the corresponding position function. Explain why this intuition suggests that the signed area under the velocity graph on an interval is equal to the difference in the position function on that interval, and tell what this has to do with the Fundamental Theorem of Calculus.

Step-by-Step Solution

Answer

The signed area under the velocity graph on an interval is equal to the difference in the position function on that interval because $s\left(t\right)$ is an anti-derivative of $v\left(t\right)$.

Step 1. Given Information.

Intuitively, the average value of a velocity function should be the same as the average rate of change of the corresponding position function.

Step 2. Explaining the intuition.

Let $v\left(t\right)$ be the velocity function from $a$ to $b$seconds and $s\left(t\right)$ be the position function.

So, $$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}v\left(t\right)dt=\frac{s\left(b\right)-s\left(a\right)}{b-a} \\ \Rightarrow {\int }_a^{b}v\left(t\right)dt=s\left(b\right)-s\left(a\right) \end{gathered}$$

The signed area under the velocity graph on an interval is equal to the difference in the position function on that interval because $s\left(t\right)$ is an anti-derivative of $v\left(t\right)$.