The derivative function for a single value of x will tell you the slope of the curve of the function at that same value of x. What does the integral function at a single value of x tell you about that same value in the original function? The only use of integrals that I’ve seen has been to compute the area under the function between two values of x.
Technically, I don’t think there is a way to evaluate an integral at a single point. The fundmamental theorems of the calculus link the relationship between an antiderivative and a definite integral, but this doesn’t mean they’re the same thing. The value of an antiderivative at a single point, as far as I know, is meaningless in terms of integral calculus, but I’ve only gone as far as CalcII in my studies, someone farther along might know more.
Also, the geometric interpretation is “area under the curve.” A physical interpretation might be “work done”, a different interpretation might be “arc length”… you know how math is, it could mean lots of things, depending on how it is used.
IIRC you can calculate the derivative function of a function, and that is it. For example the derivative function of y = x^2 is dy/dx = 2x
By contrast you can calculate the integral function of a function, but it is calculated with an unknown constant. For example the integral function of y =x^2 is (integral)y.dx = (x^3)/3 + C
So you can always calculate a value of a derivative function for any value x, but you can never calculate the integral function, only the difference of the integral function for two values x1 and x2
I should mention that you can use antiderivatives in this way. Some differential questions have a complete solution, like
dy/dx = Bx.
Such diffy-qs as linked are used for population growth, radioactive decay, Newton’s law of cooling, and other examples where the rate of change is proportional to the quantity present in a linear way.
So I guess it depends on how you read the relationship between antiderivatives and definite integrals. But in this case, you’d definitely get sensible answers from a single value. But I’m not sure this is precisely what you had in mind.
What I should have said is: “So you can always calculate a value of a derivative function for any value x, but you can never calculate the value of the integral function, only the difference of the value of the integral function for two values x1 and x2”
And what erislover said.
So here’s the deal:
There are two different differentiation operators in discussion here. One is a member of (R -> R) -> (R -> R) – that is, it maps functions from reals to reals to functions from reals to reals. That’s the indefinite derivative.
The definite derivative is a member of ((R -> R) X R) -> R. It maps a pair consisting of a function from reals to reals and a real to a real. The geometric interpretation exists, but it’s not terribly relevant here.
The indefinite integral is a member of (R -> R) -> 2[sup](R -> R)[/sup] – it maps functions from reals to reals to sets of functions from reals to reals. That’s because any two differentiable functions that differ by a constant are antiderivatives of the same function.
Now, the definite integral is a member of ((R -> R) X 2[sup]R[/sup]) -> R. This one maps a pair consisting of a function from reals to reals and a set of reals to a real number. That’s why it doesn’t make sense to ask about the integral of a function at a single value–that’s just not in the domain of the operator.
Mind you, you can take the integral of a function over a set consisting of exactly one point, but that integral will always be equal to 0.