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In the section establishing that integrals and derivatives are inverse to each other, James Stewart’s Calculus textbook says (pp325–pp326, Sec.4.3, 8Ed):

When the French mathematician Gilles de Roberval first found the area under the sine and cosine curves in 1635, this was a very challenging problem that required a great deal of ingenuity. If we didn’t have the benefit of the Fundamental Theorem of Calculus, we would have to compute a difficult limit of sums using obscure trigonometric identities. It was even more difficult for Roberval because the apparatus of limits had not been invented in 1635.

I wonder how Gilles de Roberval did it. Wikipedia and MacTutor do not contain much info on that. How to apply the method of quadrature is exactly the real challenge I suppose.

- Algebra book(s): Beginner through to advanced
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- Solve $ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $
- Equivalence of geometric and algebraic definitions of conic sections
- How to prove floor identities?

This is mainly a history question, but I’m also curious as to how one would approach this in modern days. Thank you.

P.S. I’m not sure if this is the right place for this question, and I will migrate this to the history StackExchange site if deemed so.

- slope of a line tangent through a point
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- If $f\in S_\infty$ and $\int_{\mathbb{R}}x^pf(x)d\mu=0$ for all $p\in\mathbb{N}$ then $f\equiv 0$?
- What is wrong with this funny proof that 2 = 4 using infinite exponentiation?
- Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane.
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You can see Roberval’s method in *Seventeenth-Century Indivisibles Revisited*, by Vincent Julien, pp. 192-194. Here is a link to the section in Google Books.

It’s a complicated geometric argument, and I haven’t worked through the details, but it doesn’t look to me like it’s the same as a Riemann sum. The proof seems to be based on constructing similar triangles with one infinitesimal side, within a circle.

Let us show that

$$ I(a,b)=\int_{a}^{b}\cos(x)\,dx = \sin(b)-\sin(a) \tag{1}$$

through Riemann sums. We have to compute:

$$ \lim_{n\to +\infty}\frac{b-a}{n}\sum_{k=1}^{n}\cos\left(a+\frac{(b-a)k}{n}\right) \tag{2}$$

but the RHS of $(2)$ is a telescopic sum in disguise, hence $(1)$ boils down to proving

$$ \lim_{n\to +\infty}\frac{(b-a)}{n}\,\cos\left(\frac{(b-a)+(a+b) n}{2 n}\right)\frac{\sin\left(\frac{b-a}{2}\right)}{\sin\left(\frac{b-a}{2 n}\right)}=\sin(b)-\sin(a)\tag{3}$$

which is not that difficult. This is my guess on de Roberval’s method.

Well, if you have an accurate scale handy…

- Cut out 1 square centimeter of paper and weigh it
- Draw or print the curve accurately on the same kind of paper
- Cut away the area above the curve for positive areas, and below the curve for negative areas
- Weigh the curve, then convert the mass to square centimeters using your measurement in the first step.

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