Antiderivative/Integral Notations

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I'm having some trouble understanding the notation for antiderivatives. For example, what does the big S represent and why is the antiderivative of the derivative you have to find represented as f(x)? Shouldn't it be f'(x)? Further, why is there d(x) beside the f(x)?

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By "\int" we mean the integration symbol. Or operator, if you want to be abstract. It's just a symbol with no significance if wrote alone. You need two more things: f(x) and dx. The first is the function you wish to integrate. The second is the "integration element". Its nature is not simple, it has to do with differential forms. For simplicity, you can think of it as an indespendable object.

So

\int f(x){}dx = \left\{ F(x)\left |\right \frac{dF(x)}{dx}=f(x) \right\}

Daniel.

P.S. Don't double post, please.
 
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This is probably the only question about basic calculus that cannot be answered without reference to higher-level math.

There are some false explanations, like if the integral is a sum of small rectangles, dx is the width of those rectangles. This is not at all true, unfortunately.

- Warren
 
Can you elaborate on that last part, Warren?
 
If Warren doesn't mind I'll jump in. An integral, \int_a^b f(x)dx can be defined as a limit of "Riemann sums" in which we divide the interval from a to b into many parts and approximate the area under the curve by a collection of rectangles, each having base \Delta x and height f(xi) for some xi in that interval. The "\int" symbol (my father used to call it "seahorse"!) is supposed to remind one of "Sum" for that Riemann sum.

Sometimes, in setting up an integral, it is convenient to imagine that f(xi) as "f(x)" and the [/itex]\Delta x[/itex] as "dx" so that the Riemann sum looks just like the integral you need to take. Warren's point, I think, is that while that is a convenient "mnemonic" it isn't exactly true! The integral is the limit of such sums, not the sums themselves.

Much the same thing happens when we regard \frac{dy}{dx} as a fraction- it's not true-\frac{dy}{dx} is not a fraction, but it is the limit of fractions and can be treated like a fraction.
 
Well, most students learn integrals first by studying various approximations, like the trapezoidal rule, Simpson's rule, etc. They learn that integrals are basically the sums of the areas of many small slices, and of course, the area of a slice depends upon its width.

In the limit as the slice width goes to zero, the approximation becomes an exact integral. The teachers then like to tell students that the "dx" is an infinitesimal change in x, and that it's there as the width of the infinitesimally thin rectangles.

This, of course, is not true -- on many levels. "dx" never denotes an infintisimal. It's really a one-form, a beast that maps vectors into real numbers, and it's there because you can't integrate functions directly -- that's meaningless. You can only integrate differential forms. By including a "dx," you are actually multiplying the function by a differential form, which produces a differential form, and then you can integrate it.

"dx" is the simplest such differential form, which simply denotes a line integral over the x-axis.

- Warren
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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