Example 4.2.1.
If \(h(t)\) is the rate of temperature change (Joules per second) at time \(t\) in seconds elapsed, what is the following integral?
\begin{equation*}
\int_0^7 h(t) dt
\end{equation*}
Solution.
\(h(t) dt\) is the product of rate of change and time, so it is the Joules increased. The integral adds this over the first 7 seconds, so this is the total energy (heat) increase over seven seconds. In simple terms, it is how much hotter the object is after 7 seconds.