Here, $ (dx)^2$ means $dx \wedge dx$, and the fact that it vanishes comes from the fact that the exterior algebra is anti-commutative. In other words, formally we have $d^2x=0$ and $ (dx)^2=0$ but …

A "signed definite integral" for computing work and other "net change" calculations. The value of an expression such as $\int_0^1 x^2\,dx$ comes out the same under all these interpretations, of course. …

I am working on trying to solve this problem: Prove: $\\int \\sin^n{x} \\ dx = -\\frac{1}{n} \\cos{x} \\cdot \\sin^{n - 1}{x} + \\frac{n - 1}{n} \\int \\sin^{n - 2}{x ...

I know dy/dx for example means "derivative of y with respect to x," but there's another context that confuses me. You will generally just see a dx term sitting at the end of an integral equation an...

How to integrate $\int \frac {1} {\sin^4x + \cos^4 x} \,dx$? Ask Question Asked 11 years, 6 months ago Modified 3 years, 2 months ago

I just finished taking my first year of calculus in college and I passed with an A. I don't think, however, that I ever really understood the entire $\\frac{dy}{dx}$ notation (so I just focused on ...

The symbol used for integration, $\int$, is in fact just a stylized "S" for "sum"; The classical definition of the definite integral is $\int_a^b f (x) dx = \lim_ {\Delta x \to 0} \sum_ {x=a}^ {b} f (x)\Delta x$; the limit …

This allows us to freely choose any expression for $dx$ and by definition we will get the exact expression for $dy$ so that $dy/dx$ (the fraction) is the same as the single symbol $dy/dx$ (the …

For a definite integral with a variable upper limit of integration $\int_a^xf (t)\,dt$, you have $ {d\over dx} \int_a^xf (t)\,dt=f (x)$. For an integral of the form $$\tag {1}\int_a^ {g (x)} f (t)\,dt,$$ you would find the …

May 2, 2015 · Okay this may sound stupid but I need a little help... What do $\Large \frac {d} {dx}$ and $\Large \frac {dy} {dx}$ mean? I need a thorough explanation. Thanks.