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

elvishwitchxyp

elvishwitchxyp

Answered question

2022-01-03

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

Answer & Explanation

Bubich13

Bubich13

Beginner2022-01-04Added 36 answers

This is a generalisation of the Riemann integral:
Before we had:
abdx=xab
Now we have
abd(g(x))=g(x)ab
abd(g(x)) is somewhat like abg(x)dx, much like in probability:
If we have a continuous random variable X in (mathscr{F},P), then
P(Xx)=xdFX(x)
If X has a pdf, then we have
P(Xx)=xFx(x)dx=xfX(x)dx
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)]=Rg(x)FX(x)dx=Rg(x)fX(x)dx
If X has a pdf, then we have
E[g(X)]=Rg(x)FX(x)dx=Rg(x)fX(x)dx
lovagwb

lovagwb

Beginner2022-01-05Added 50 answers

When f' 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

karton

Expert2022-01-09Added 613 answers

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

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