elvishwitchxyp

2022-01-03

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

Bubich13

This is a generalisation of the Riemann integral:
${\int }_{a}^{b}dx=x{\mid }_{a}^{b}$
Now we have
${\int }_{a}^{b}d\left(g\left(x\right)\right)=g\left(x\right){\mid }_{a}^{b}$
${\int }_{a}^{b}d\left(g\left(x\right)\right)$ is somewhat like ${\int }_{a}^{b}{g}^{\prime }\left(x\right)dx$, much like in probability:
If we have a continuous random variable X in ($mathscr\left\{F\right\},\mathbb{P}$), then
$P\left(X\le x\right)={\int }_{-\mathrm{\infty }}^{x}d{F}_{X}\left(x\right)$
If X has a pdf, then we have
$P\left(X\le x\right)={\int }_{-\mathrm{\infty }}^{x}{F}_{x}^{\prime }\left(x\right)dx={\int }_{-\mathrm{\infty }}^{x}{f}_{X}\left(x\right)dx$
Not all continuous random variables have pdfs so we can use
Same with expected value: Let g be a Borel-measurable function. Then
$E\left[g\left(X\right)\right]={\int }_{\mathbb{R}}g\left(x\right){F}_{X}^{\prime }\left(x\right)dx={\int }_{\mathbb{R}}g\left(x\right){f}_{X}\left(x\right)dx$
If X has a pdf, then we have
$E\left[g\left(X\right)\right]={\int }_{\mathbb{R}}g\left(x\right){F}_{X}^{\prime }\left(x\right)dx={\int }_{\mathbb{R}}g\left(x\right){f}_{X}\left(x\right)dx$

lovagwb

When f' is continuous on $\left[a,b\right],$
${\int }_{a}^{b}df\left(x\right)={\int }_{a}^{b}{f}^{\prime }\left(x\right)dx=f\left(b\right)-f\left(a\right)$

karton

Owing to the uncertain integrals property
$\int dF\left(x\right)=F\left(x\right)+const$

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