B Question about units for "area under curve"

[Chenkel]

Hello everyone,

I am curious, suppose you have a function $f(x)=x^3$ and you to find the area under the curve from 0 to x, the area would be $\frac {x^4}{4}$ but this is units of $L^4$ if x is length, but area is units of $L^2$ so what is going on here?

The reason I'm curious is I imagine people could find applications of using complex curves in engineering to make different things and calculate area or volumes, but just because you have an interesting curve doesn't mean the units of area shouldn't check out as $L^2$ do you see what I'm saying?

Any advice is appreciated!

Thank you, and let me know what you think!

 

[Ibix]

The dimension of the area under a curve depends what units you specify on the axes. Take, for example, a plot of velocity versus time. The integral is the displacement, because the height of a strip has dimension $[LT^{-1}]$ and the width has dimension $[T]$.

In your case, you've specified that the vertical axis has dimension $[L^3]$ and the horizontal has dimension $[L]$, so the area has dimension $[L^4]$. If you want $f$ to have dimension $[L]$ you need to do something like define it as $f(x)=a^{-2}x^3$ where $a$ is some constant with dimension $[L]$.

 

[hutchphd]

I applaud your attention to units. "Area under the curve" refers to the area on the graph paper that represents the function. Each square of the graph paper then represents a unit of dimension [x f(x)] and the total area will represent the physical value of the desired integral. Before digital computers folks sometimes did integrals by cutting out the curve trace on paper and weighing it. Then the useful number was the "weight" of the curve (not the area)......same idea..