What are the units of a definite integral and its derivative?

[AbsoluteZer0]

Hi,

Suppose we have f(x) = x³.
Integrating this function using the definite integral with the upper boundary being 3 and the lower boundary being 1 would result in 20. Does 20 have any units or is it unitless? Seeing how it is the area underneath a curve, I would imagine that it has square units. If you were to differentiate this function you would get \frac{dy}{dx}=3x². Does the derivative have any units?

Thanks,

 

[Vargo]

Say y=f(x).

When you integrate the units are the product of the units of x times the units of y.

If you differentiate dy/dx the units are the units of y divided by the units of x.

Ex: If y has length units (m) and x has time units (s), then the integral will have units m*s and the derivative will have units m/s.

If y=x^3, then your units for y might be the units of x cubed, but it depends on the problem. If y = volume of a cube with sidelength x, then units of y are units of x cubed.

 

[HallsofIvy]

Recall that a integral is a limit of a sum,the Riemann sum, \sum f(x_i)\Delta x so its units are that of a product of whatever units f(x) has and whatever units x has. That is why we typically introduce integrals in terms of "area under the curve"- if x has units of, say meters, and y= f(x) has units of meters also, then the integral \int f(x)dx has units of "meters squared" or "square meters", an area measure.

On the other hand, if v(t) is a velocity, so that v has units of "meters per second" and t has units of "seconds" then the integral has units of "meters per second" times "seconds" or "meters": \int v(x)dx will give the distance the object has moved.

The derivative is the other way around. It is a limit of \Delta y/\Delta x so it has units of what ever the units y has divided by the units of x. If f(t) is the distance, in meters, an object moves in t seconds, then df/dt has units of "meters per second", a speed.