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I just started learning "rigorous" calculus, and I noticed that a lot of calculus theorems are rather obvious from the geometrical point of few.

Some examples:

1. Prove that the derivative of an odd (resp. even) function, when exists, is even (resp. odd)

Since the graph is symmetric with respect to the origin, the slopes of the tangent lines at $x=a$ and $x=-a$ must equal, hence the derivative is even. Similarly for the even function.

2. If $f$ is one-one and continuous, then $f^{-1}$ is also continuous.

This can't be more obvious. $f^{-1}$ is just the reflection of $f$ about the line $y=x$.

3. $\displaystyle\int_a^bf(x)dx+\int_b^cf(x)dx=\int_a^cf(x)dx$

The sum of the area from $a$ to $b$ and the area from $b$ to $c$ is, of course, the area from $a$ to $c$.

4. If $f$ is a one-one function one $[a,b]$, then $\displaystyle\int_a^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a).$

Quite clear if you draw a diagram. The sum of the two integrals is the difference of two rectangles.

5. If $f$ is increasing and $f(0)=0$, then for $a,b>0$ we have $\displaystyle\int_0^af(a)dx+\int_0^bf^{-1}(x)dx\ge ab$.

Also clear from the diagram. There's a "leftover" part outside the rectangle of area $ab$.

Of course, these proofs are not rigorous, and possibly not valid at all. Firstly, I only consider the easy cases. Secondly, I used the intuitive properties of geometric objects without proofs.

So I'm wondering, is there any theory that connects calculus with geometry rigorously? Such a theory would help simplify calculus proofs tremendously, as I roughly outlined above.

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In the same spirit as #3: consider the integral of an odd or even function over a symmetric interval... –  J. M. Oct 15 '11 at 5:37
    
Thanks, but I'm not interested in more examples. My question is, how can we make those proofs rigorous? Is there any theory that connects calculus with geometry? –  123 Oct 15 '11 at 6:02
    
In order to make the "geometric" proofs rigorous themselves, you'd have to actually be more explicit and describe the "picture" that runs through your head. This would be the only way to prove things. More often than not, the proofs themselves are derived from a well-drawn picture, so I'm not sure what you're getting at. Also, to address your second question, I am assuming you're not talking about extending calculus to non-Euclidean spaces, such as things that locally look like $\mathbb{R}^n$, i.e. manifolds. –  Landyn Oct 15 '11 at 6:12
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Because of the fallibility of geometric intuition, It has been standard for at least 150 years to "arithmetize" geometry rather than the other way around. Continuous functions can be a lot weirder than anything any pencil imagined drawing. –  André Nicolas Oct 15 '11 at 6:47
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Geometric intuition is essential to get an idea of what might be true, and how we might try to prove it. –  André Nicolas Oct 15 '11 at 7:06
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1 Answer

As people have already told you in the comments, geometric interpretations of problems are extremely useful to find the solution, but then to make it rigorous you have to be very careful and prove things formally. The reason why you have to do it is fallibility of geometric intuition as @Andre wrote. Maybe I will not answer completely your question, but there are three examples about geometry and they are not the only:

  1. Fat Cantor Set: could you imagine that the subset of $[0,1]$ with an empty interior, dense nowhere can have a length very close to $1$?

  2. Poincare conjecture formulation may seem to be very logical at the first glance, simply solvable by geometric methods you described though this problem appeared to be complicated and offered a money prize comparable with Nobel prize.

  3. Finally, have you read about Banach-Tarski paradox? That fact seems to be impossible if you only rely on simple geometrical arguments.

One more point: if you're interested in more motivation and examples, these books can be of interest for you: for calculus see Counterexamples in Analysis and for more complicated but not less geometrical math see Counterexamples in Topology.

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