Is infinitesimal an unbounded variable? Please explain Or bounded variables

Infinitesimal is neither an unbounded variable nor a bounded variable. Infinitesimal is a very small number, close to zero, while unbounded variable is bounded and unbounded (tends to infinity), and bounded variable is bounded on both sides

What is arctan (arctanx) equal to?

It's equal to X

What is the second mean value theorem for integrals?

He's wrong. I'm sure he's right. We've learned,
Let f (x) be integrable on [a, b], and G (x) be monotone on [a, b],
Then there exists ξ ∈ [a, b], such that
∫abf(x)g(x)dx=g(a)∫aξf(x)dx+g(b)∫bξf(x)dx
A is the lower bound of the integral, B is the upper bound of the integral,
After that, a is the lower bound, ξ is the upper bound, then ξ is the lower bound, and B is the upper bound of the integral,
It's so painful to play the formula

The proof process of the second mean value theorem of integral
I know the content of the second mean value theorem of integral. I want to know how to prove the second mean value theorem of integral?

Second integral mean value theorem second integral mean value theorem:
If 1) f (x) is nonnegative decreasing on [a, b],
(2) G (x) is integrable on [a, b],
Then there is an open interval (a, b) where the integral value of F (x) g (x) in [a, b] is equal to the integral value of F (a + 0) times g (x) in [a, C]
inference
If (1) f (x) is monotone in [a, b],
(2) G (x) is integrable in [a, b],
Then there exists an open interval (a, b) where the integral value of F (x) g (x) in [a, b] is equal to the sum of the integral value of F (a + 0) multiplied by G (x) in [a, C] and the integral value of F (B-0) multiplied by G (x) in [C, b]

What are the contents of the first mean value theorem and the second mean value theorem

First, if f (x) is continuous on [a, b], then there exists at least one point ξ on [a, b], such that ∫ (a, b) f (x) DX = f (ξ) (B - a). Second, Let f (x) be integrable on [a, b] and G (x) monotone on [a, b], then there exists ξ ∈ [a, b], such that ∫ (a, b) f (x) g (x) DX = g (a) ∫ (a, ξ) f (x) DX + G (b) ∫ (...)

How to prove the extension of the mean value theorem of integral

First, if the functions f (x) and G (x) are integrable in the closed interval [a, b], and f (x) is a monotone function, then there is at least one point ξ in the integral interval [a, b], which makes the following formula true:
This can be done with the zero point theorem

Some problems in the proof of integral mean value theorem
The function is continuously differentiable in [0,1], and 3 times the above integral 1, the lower integral 2 / 3 F (x) DX = f (0) has at least one point C in (0,1), so that f '(c) = 0

The integral mean value theorem shows that there is a point x0,2 / 3

Why is ln (1 + x) approximately equal to X / (1 + x) when x is far less than 1

When x approaches 0, DF (x0) = f (x0 + x) - f (x) = f (x0)'x, so f (x0 + x) = f (x0) + F '(x0) x, let x0 = 0, then f (x) = f (0) + f' (0) X

X tends to 0, find ln (1 + 2x) / X

X tends to zero, so 2x tends to zero
So ln (1 + 2x) and 2x are equivalent infinitesimals
So the original formula = LIM (x tends to 0) (2x / x) = 2

What does limx tend to be 0 ln (1 + 2x) / (2 ^ x-1)

The first derivative of numerator denominator is obtained by using first order Robita
lim[x→0]ln(1+2x)/(2^x-1)
=lim[x→0][2/（1+2x）]/（2^xln2）
=2/ln2
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Is infinitesimal an unbounded variable? Please explain Or bounded variables