User:LavosBacons/LaTeX

$\int _{1}^{t}\pi f(x)^{2}dx=(1/6)\pi [(1+3t)^{2}-16]$

$\int _{1}^{t}f(x)^{2}dx=(1/6)[(1+3t)^{2}-16]$

Let $g(x)=f(x)^{2}$ and let $G(x)=$ an antiderivative of $g(x)$. Then:

$\int _{1}^{t}g(x)dx=(1/6)[(1+3t)^{2}-16]$

$G(t)-G(1)=(1/6)[(1+3t)^{2}-16]$

$G(t)=(1/6)[(1+3t)^{2}-16]+G(1)$

${\frac {d}{dt}}G(t)={\frac {d}{dt}}((1/6)[(1+3t)^{2}-16]+G(1))$

${\frac {d}{dt}}G(t)=(1/6){\frac {d}{dt}}[(1+3t)^{2}]$

${\frac {d}{dt}}G(t)=(1/6){\frac {d}{dt}}[1+6t+9t^{2}]$

${\frac {d}{dt}}G(t)=(1/6)[6+18t]$

${\frac {d}{dt}}G(t)=1+3t$

$g(t)=1+3t$

Plugging back in,

$g(x)=f(x)^{2}$

$1+3x=f(x)^{2}$

$f(x)={\sqrt {1+3x}}$

$marily_{n}=marily_{n-1}+marily_{n-2}$

${\sqrt {2}}$

$\Sigma _{i=1}^{\infty }10^{-(i!)}$

$\neg \exists x\ \exists L\ \exists \ a>0\ \forall \ \epsilon \in (0,a)\ \exists \ \delta \ |love\ for\ you\ (x+\delta )-L|<\epsilon$

${\frac {d}{dx}}\int f(x)dx=f(x)$