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Calculus I: Activities

Section 4.2 Interpreting Integrals

Standards
  • Demonstrate understanding of parts of definition of integral using an application
Our initial motivation for integrals in Subsection 4.1.1 was area. However, there are many applications which have the same structure.

Subsection 4.2.1 Interpret an Integral

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.

Example 4.2.2.

If \(h(t)\) is the rate of temperature change (Joules per second) at time \(t\) in seconds elapsed, what is each part of the integral defintion?
\begin{equation*} \int_0^7 h(t) dt = \lim_{|\Delta x| \to 0} \sum_{i=0}^n h(t_i)(t_{i+1}-t_i) \end{equation*}
Solution.
  • \(h(t_i)\) is the rate of heat transfer at time \(t_i\)
  • \((t_{i+1}-t_i)\) is the length of the time interval
  • \(h(t_i)(t_{i+1}-t_i)\) is the average energy increase in that time interval.
  • \(\sum_{i=0}^n\) sums the average energy absorbed for an approximate total heat increase.
  • \(|\Delta x|\) is the maximum length time interval.
  • \(|\Delta x| \to 0\) means the maximum time interval is shrinking to zero.
  • 0 is the start time.
  • 7 is the end time.
  • The integral calculates the increase in temperature over 7 seconds.