What is the relation of infimalistic math to calculus?

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in my previous course i solved(with your help) every type of calculus question
without hearing or reading on infimalistic math.

now i am doing this course
and i talked to the lecturer and he told me that
in this course i have the normal questions and proofs
but each proof question comes from each type of question that i already know.

so why do i need infimalistic math?
 
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What is "infimalistic math"??
 
"Infimalistic"? Do you mean "infinitesmal"? You may be referring to "non-standard" analysis which begins by extending the real number system to include both "infinite numbers" and "infinitesmal numbers". That allows one to say, for example, that if y= x2, then y+ dy= (x+ dx)2= x2+ 2xdx+ (dx)2 so dy= 2xdx+ dx2 and dy/dx= 2x+ dx. By the laws of arithmetic that have to be introduced to deal with "infinitesmal" and "infinite" numbers, dy/dx, the ratio of two "infinitesmal" numbers is a "regular" number while on the right, the sum 2x+ dx, of a "regular" number and a while dx is "infinitesmal" so their sum is just 2x: dy/dx= 2x.

There are no results in "non-standard" analysis that are not true in "standard" analysis so you don't need to know "non-standard" analysis for applications.
 
the are therms like SUP and INF which i am required to know in this course
is that part of "non-standard" analysis("infinitesmal math")?
 
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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