Life Has Intrinsic Value: A Graphic Proposition

A little over a month ago I was pressed to prove or justify why I believed life has intrinsic value (valuable in itself, for its own sake). With this article


Integral from 0 to ∞ [x^a/(x^b+1)] dx

\begin{align*}& I(a, b) = \int_0^\infty \dfrac{x^a}{x^b+1} dx =\dfrac{\beta(\dfrac{a+1}{b}, 1-\dfrac{a+1}{b})}{b} \\& \\& {\rm{Proof}}: \\& \\& \therefore u=x^b \Longrightarrow \int_0^\infty \dfrac{u^{\frac{a-b+1}{b}}}{b(u+1)} du \\& \\& \therefore \int_0^\infty \dfrac{u^{\frac{a-b+1}{b}}}{b(u+1)} du = \dfrac{\beta(\dfrac{a+1}{b}, 1-\dfrac{a+1}{b})}{b} \\&

Integrals: Royal Properties

\begin{align*}& {\rm{Given}}:\int_0^\infty f(x) dx \\& \\& (1) \: \rm{Queen \: Property} \\& \\&\therefore u=(x)^{-1} \Longrightarrow\int_0^\infty f[(x)^{-1}] (x)^{-2} dx \\& \\& (2) \: \rm{Jacksonian \: Property} \\& \\&\therefore u=(x)^{-n} \Longrightarrown \int_0^\infty

Integral of [1+(x^g)]^(-n) from 0 to ∞

\begin{align*}& \int_0^\infty \dfrac{1}{(1+x^g)^n} dx =\int_0^\infty \dfrac{(x)^{gn-2}}{(1+x^g)^n} dx \\& \\& \therefore x^g = t \: , \: x = t^{\frac{1}{g}} \Longrightarrowdx = \dfrac{t^{\frac{1}{g}-1}}{g} dt \\& \\& \therefore \int_0^\infty \dfrac{(x)^{gn-2}}{(1+x^g)^n} dx \Longrightarrow\int_0^\infty

Integral of (1+[x^2])^(-n) from 0 to ∞

\begin{align*} & \int_0^{\infty} \dfrac{1}{(1+x^2)^n} dx \\& \\& \\& {\rm{Lemma \: (1)}}: \int_0^{\infty} \left[ f(x) \right] dx = \int_0^{\infty} \left[ \dfrac{f(\frac{1}{x})}{(x^2)} \right] dx \\& \\& {\rm{Lemma \: (2)}}: \int_0^{\frac{\pi}{2}} \left[\sin(x)\right]^{2g-1} \left[\cos(x)\right]^{2h-1}

200m MLR Estimate

16 April 2019 Formula (19/718 sampled) \begin{align*} \rm{200m} &\approx \rm{100m} \left[ 1.997383 + \left( \dfrac{7.633557}{\rm{60m}} \right) \right] - 11.1841 \pm 0.3 \\& \approx\rm{100m} \left[ 2 + \left( \dfrac{7.65}{\rm{60m}} \right) \right]

I'm an INTJ - A

Sid ... INTJ's (referred to as "Architects" and "The Master Mind") are a very rare personality type. They constitute ~ 2.1% of the populace (in regards to males ~3.3%). I

Integral of [a(x^n)+b]/[c(x)+d]

\begin{align*}I = \displaystyle \int_\alpha^\beta\dfrac{a(x)^n+b}{cx+d} \: dx &= \displaystyle \dfrac{a}{c} \int_\alpha^\beta\dfrac{(x)^n+ \frac{b}{a}}{x+\frac{d}{c}} \: dx \\&= \displaystyle \dfrac{a}{c} \int_\alpha^\beta\dfrac{(x)^n+ \Delta_1}{x+\Delta_2} \: dx \\& \therefore \: u=x+\Delta_2 \: \Longrightarrow \: u\mid_{\alpha+\Delta_2=\Delta_3}^{\beta+\Delta_2=\Delta_4} \\& \therefore

Integral of 1/(x^n+1) from 0 to ∞

\begin{aligned} I = \int_0^\infty \dfrac{1}{(x)^n+1} dx  = \dfrac{\pi}{n\sin{( \frac{\pi}{n}} )} \end{aligned} \begin{aligned} \therefore \rho=\left[ (x)^n+1 \right]^{-1} \:;\: \rho \mid_0^1 \: \Longrightarrow d\rho=\dfrac{-n(x)^{n-1}}{\left[ (x)^n+1 \right]^2} \: dx\end{aligned} \begin{aligned} \therefore (n)^{-1} \int_0^1 \left[

Integral of 1/[(x^3)+1] from 0 to ∞

I=∫∞01x3+1dx=2π31.5≈1.209199576156145 Partial Fraction Decomposition Method \begin{aligned} \int_0^{\infty} \dfrac{1}{(x+1)(x^2-x+1)} \,dx \end{aligned}.\begin{aligned} \int_0^{\infty} \dfrac{1}{3(x+1)} \,dx+ \int_0^{\infty} \dfrac{x-2}{3(x^2-x+1)} \,dx \end{aligned}.\begin{aligned} \dfrac{1}{3} \int_0^{\infty} \dfrac{1}{x+1} \,dx-\dfrac{1}{3} \int_0^{\infty} \dfrac{x-2}{x^2-x+1} \,dx \end{aligned}.\begin{aligned} \left[\dfrac{1}{3} \ln{(x+1)}\right]\mid_0^{\infty}-\dfrac{1}{6}\int_0^{\infty} \dfrac{2x-1}{x^2-x+1} \,dx+\dfrac{1}{2}\int_0^{\infty} \dfrac{1}{x^2-x+1}

Fitness Plans

Below comprises 11 fitness programs/plans I offer. Once you've purchased a plan expect an intial fitness goal and evaluation via email. - C. D. Chester All Products General Nutrition /

Other Personal Outlets

So I also publish on Medium. You can see my articles as stories over there. Here's the link: C. D. Chester's Latest Medium Posts Everything I post on this website starting

VO2 Max and its Relations

References: VO2 max by Brian Mackenzie Estimation of  VO2max from the ratio between HRmax and HRrest – the Heart Rate Ratio Method by Uth et al VO2 max from a one mile