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Units of a definite integral

  1. Nov 13, 2012 #1

    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 [itex]\frac{dy}{dx}[/itex]=3x2. Does the derivative have any units?

  2. jcsd
  3. Nov 13, 2012 #2
    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.
  4. Nov 13, 2012 #3


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    Recall that a integral is a limit of a sum,the Riemann sum, [itex]\sum f(x_i)\Delta x[/itex] 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 [itex]\int f(x)dx[/itex] 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": [itex]\int v(x)dx[/itex] will give the distance the object has moved.

    The derivative is the other way around. It is a limit of [itex]\Delta y/\Delta x[/itex] 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.
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