Just a silly question about calculus

[1MileCrash]

In problems regarding rate of change, like velocity can be related to the derivative, while problems regarding the rate of the rate of change, acceleration, can be related to the second derivative.

Definite integrals are related to area, are there any applications, or even a such thing as a higher order integral, the integral of an integral?

I always assumed volume before I learned, but after learning those problems, it isn't the case. So what does a integral of an integral represent to the original function?

 

[pmsrw3]

Yes, there are double, triple, etc integrals, although typically they aren't over the same variable. In fact, the integral of an integral can be volume.

I think, however, that it isn't really all that helpful to think of the integral as area. I'm not saying it's wrong -- for some problems, an integral will indeed give you area. But a more general idea is that an integral is a sum -- it adds things up. So, if you integrate acceleration twice to get position, the first integral is adding up all the acceleration to give velocity, and the second integral is adding up all the velocities to get distance.

 

[1MileCrash]

I can visualize the adding up, like a cup filling up with water from the function to the integral as the cup moves along the x axis. That's my mind movie.

So what you're saying is that it's explicitly the inverse of differentiation, that if I integrate a constant speed, 5, finding the definite integral over a closed set shows the total distance traveled from the start to finish of that set. Right? From [1,2] the definite integral is 5, of course, from [1,1] it is zero, from [1,3] it is 10.

I really doubt most people doing integration problems have ever thought about that.

 

[pmsrw3]

So what you're saying is that it's explicitly the inverse of differentiation, that if I integrate a constant speed, 5, finding the definite integral over a closed set shows the total distance traveled from the start to finish of that set. Right? From [1,2] the definite integral is 5, of course, from [1,1] it is zero, from [1,3] it is 10.

Yup, that's exactly right. And the cool thing is, it works even if the speed isn't constant.

 

[1MileCrash]

I'd sure hope so! :)

 

[HallsofIvy]

A very common application is this: if the acceleration of some object is a(t) then its speed is given by $v(t)= \int a(t)dt$ and then the distance moved by $\int v(t) dt$. Of course, that is not a perfect "inverse". Since the derivative of any constant is 0, the derivative is not "one-to-one" and so its "inverse", integration, is not well defined- we need to add some arbitrary constant. If, for example, the acceleration of an object is the constant a(t)= -9.81, then its speed, at any instant t, is [itex]v(t)= \int -9.81 dt= -9.81t+ C[/tex]. Of course, taking t= 0, v(0)= C so C is the intial velocity: v(t)= -9.81t+ v(0). We get the postion, at any instant t, by integrating again:<br /> $x(t)= \int (-9.81t+ v(0))dt= -4.905t^2+ v(0)t+ C$ and it is easy to see that this constant, C, is the initial position, x(0). That is, by the "second integral" of a(t)= -9.81, $x(t)= -4.905t^2+ v(0)t+ x(0)$.[/itex]