Visual calculus

I've been wondering: is it possible to visualize the rules of calculus?

To simplify things, in the above visualization

$v(x_1) = 0$ , so that

$\int^{x'}_{x_1}\frac{dv}{dx}dx = v(x')$ . This means that

$v(x_2)$ is simply the area of the cross-section facing us on the LHS, and the volume on the LHS is

$u(x_2)v(x_2)$ .

The RHS shows the same volume split into two parts. The rectangle embedded in the first is

$u\frac{dv}{dx}$ , and so its volume is

$\int^{x_2}_{x_1}u\frac{dv}{dx}dx$ . The cross section of the 2nd part is

$v(x)$ , so its volume is

$\int^{u_2}_{u_1}vdu$ , which becomes

$\int^{x_2}_{x_1}v\frac{du}{dx}dx$ when rewritten as an integral over

$x$ .

So the picture is visual proof that the equation in pink holds, provided that

$v(x_1) = 0$ . To prove it in general we just need to check that when we replace

$v$ with

$v+v_1$ in the equation, both sides change by the same amount.

This picture demonstrates the chain rule for differentiation.

$z$ is a function of

$y$ which is a function of

$x$ . If

$x$ changes by

$\delta x$ then

$y$ changes by, approximately,

$\frac{dy}{dx}\delta x$ , and so

$z$ changes by, approximately,

$\frac{dz}{dy}\frac{dy}{dx}\delta x$ .

$u$ changes by

$\delta u$ and

$v$ by

$\delta v$ then

$uv$ changes by

$v\delta u + u\delta v + \delta u\delta v$ . Dividing this by

$\delta x$ and taking the limit gives the result.

Note that if you integrate both sides of the product rule you get the integration by parts rule.

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