I Integrals are harder than derivatives, why?

[cosmik debris]

I didn't really understand these subtleties until a couple years after my first calculus sequence. Applications in physics helped, but what helped the most was when I finally got to a numerical analysis course and learned how to compute just about any derivative or integral numerically (and also how to integrate differential equations numerically. Somehow the computational approach (as opposed to the pencil and paper analytical approach) was the missing piece for my conceptual understanding.

Yes, we found that showing high school students how to write a program for integrating and differentiating gave them a much better understanding than the analytical approach.

Cheers

 

[haruspex]

or is just another way of saying it,

Yes, I feel your post neatly illustrates the sense in which integration is harder, but doesn't really constitute an explanation.
E.g. why are there not equivalents to the product rule and chain rule? (As Mark44 notes in post #9, integration by parts is equivalent to the product rule for differentiation, so is not an integration analogue of it.)
What would it look like?
Consider functions x=x(z) and y=y(z). Integrating up from z=0, ∫₀^cx.dz can be visualised as an area in the XZ plane between z=0 and z=c. Likewise, ∫₀^cy.dz in the YZ plane.
∫₀^cxy.dz, though, is a volume with those two shapes as faces and rectangular cross-section at each z.
Clearly we cannot deduce the volume merely from the two areas - it depends on the detailed interplay between x(z) and y(z) over the range of z. So it is not possible to write it in terms of x(c), y(c), ∫^cx and ∫^cy.

 

[Demystifier]

Is there a way to define the concept of indefinite integral without referring to the concept of a derivative? If no, or if it cannot be done in a simple way, then it could be the reason why analytical computation of indefinite integrals is harder than that of derivatives.

By the way, I remember that there is an advanced textbook on mathematical analysis that teaches integration before derivatives, but I cannot recall which textbook is that. Can someone refresh my memory?

 

[Demystifier]

Or perhaps the reason is the following. Suppose that function $f(x)$ is given for all $x$. From this, we want to determine the derivative $f'(x)$ and the anti-derivative $F(x)$. But instead of determining it at all points $x$, let us concentrate on one particular point $x_0$, say $x_0=7$. So now we only want to determine the two numbers $f'(x_0)$ and $F(x_0)$. The crucial difference is that the task of finding $f'(x_0)$ is well defined, while the task of finding $F(x_0)$ is meaningless. With fixed $f(x)$, the number $F(x_0)$ can be any number. Intuitively, this means that $f'(x)$ is a local property of a function $f(x)$, while $F(x)$ is a non-local property of a function $f(x)$. The $F$ only makes sense if it is defined in a finite neighborhood of a point $x_0$, while $f'$ can be defined at $x_0$ without knowing it in its neighborhood. Intuitively, determining something in a whole neighborhood should be more complicated than determining something at a single point.

 

[Demystifier]

By the way, I remember that there is an advanced textbook on mathematical analysis that teaches integration before derivatives, but I cannot recall which textbook is that. Can someone refresh my memory?

I found it!
https://www.amazon.com/dp/0387940014/?tag=pfamazon01-20

And for the record, I don't longer think that my argument in #54 is correct, but thanks for the likes!

 

[Svein]

The salient questions are:

1. If f(x) is integrable, does \frac{d}{dx}(\int f(x) dx) exist, and is it equal to f(x) a. e.?
2. If f(x) is differentiable on an open interval (a, b), will \int_{a}^{x} (\frac{d}{dt} f(t))dt be equal to f(x) - f(a)?

Neither of these are trivial...

 

[OwlHoot]

Besides a lot of waffle about formalism and terminology, I'd suggest one answer to the OP's question is that differentiation is a "local" operation at a point whereas integration is a more "global" operation across a range of points. It isn't a foregone conclusion that the first must be simpler than the second, and I can think of examples where the opposite may be true. But that may be a suitable starting point for a type of answer that will satisfy the OP more than they seem to have been so far.

That aside, I think the best answer (already alluded to above) is integration's absence of a recursive procedure as effective as using the chain rule for differentiation, although integration by parts helps up to a point.

 

[e-pie]

An Analogy

Cutting a piece of paper into infinite strips is easy. Scissors make life easy. But joining them is hard. Glue sticks.

Same goes for integration and differentiation.

When you are differentiating a function, you are drawing tangents over the curve. Almost every curve has tangents but not all curve forms a closed area to integrate. Even if it forms it goes in the hyperbolic/complex domain. But differentiating real valued functions never gives a complex result. Because the tangents are all real line segments.