Why is the integral of d1+x2 simply equal to 1+x2 ?

Question

Bubich13 · Accepted Answer

This is a generalisation of the Riemann integral:
Before we had:

\int_{a}^{b} d x = x ∣_{a}^{b}

Now we have

\int_{a}^{b} d (g (x)) = g (x) ∣_{a}^{b}

\int_{a}^{b} d (g (x))

is somewhat like

\int_{a}^{b} g^{'} (x) d x

, much like in probability:
If we have a continuous random variable X in (

m a t h s c r {F}, P

), then

P (X \leq x) = \int_{- \infty}^{x} d F_{X} (x)

If X has a pdf, then we have

P (X \leq x) = \int_{- \infty}^{x} F_{x}^{'} (x) d x = \int_{- \infty}^{x} f_{X} (x) d x

Not all continuous random variables have pdfs so we can use
Same with expected value: Let g be a Borel-measurable function. Then

E [g (X)] = \int_{R} g (x) F_{X}^{'} (x) d x = \int_{R} g (x) f_{X} (x) d x

If X has a pdf, then we have

E [g (X)] = \int_{R} g (x) F_{X}^{'} (x) d x = \int_{R} g (x) f_{X} (x) d x

lovagwb · Accepted Answer

When f&#039; is continuous on [a,b],∫abdf(x)=∫abf′(x)dx=f(b)−f(a)See more information about the relationship between Riemann Stieltjes integrals and Riemann integrals here.

karton · Accepted Answer

Owing to the uncertain integrals property ∫dF(x)=F(x)+const

Why is the integral of d\sqrt{1+x^2} simply equal to \sqrt{1+x^2}

Answered question

Answer & Explanation

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