MA005: Calculus I

We have already seen that the "area" under a graph can represent quantities whose units are not the usual geometric units of square meters or square feet. For example, if $x$ is a measure of time in "seconds" and $\mathrm{f}(x)$ is a velocity with units "feet/second", then $\Delta \mathrm{x}$ has the units "seconds" and  $\mathrm{f}(x) \cdot \Delta x$ has the units ("feet/second")("seconds") = "feet," a measure of distance. Since each Riemann sum $\sum \mathrm{f}(x) \cdot \Delta x$ is a sum of "feet" and the definite integral is the limit of the Riemann sums, the definite integral, has the same units, "feet".

If the units of  $\mathrm{f}(x)$ are "square feet" and the units of $x$ are "feet", then $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} x$ is a number with the units ("square feet"). ("feet") = "cubic feet," a measure of volume. If $f(x)$ is a force in grams, and $x$ is a distance in centimeters, then $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} x$ is a number with the units "gram. centimeters," a measure of work.

In general, the units for the definite integral $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} x$ are (units for $f(x)$).(units for $x$). A quick check of the units can help avoid errors in setting up an applied problem.