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. In more general settings, the three interpretations generalize in different ways, so that the "dx" comes to mean different things.
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...
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 of the Riemann sum of f (x) between a and b as the increment of X approaches zero (and thus the number of rectangles approaches infinity).
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.
Well, $\delta x$ means different things depending on the context. For example, it has a particular meaning in variational calculus, and a completely different one in functional calculus...
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 ...
In Leibniz notation, the 2nd derivative is written as $$\dfrac {\mathrm d^2y} {\mathrm dx^2}\ ?$$ Why is the location of the $2$ in different places in the $\mathrm dy/\mathrm dx$ terms?
I understand the meaning of $\frac {dy} {dx}$ and $\int f (x)dx$, but outside of that what do $dy, du, dx$ etc.. mean? When I took calc I, derivatives and integrals were given a definition, but these things were kind of skipped over.
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 ...
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 for two different reasons.
