User:LavosBacons/LaTeX

$\int_1^t \pi f(x)^2dx = (1/6)\pi [(1+3t)^2 - 16]$

$\int_1^t f(x)^2dx = (1/6)[(1+3t)^2 - 16]$

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

$\int_1^t g(x)dx = (1/6)[(1+3t)^2 - 16]$

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

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

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

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

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

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

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

$g (t) = 1 + 3 t$

Plugging back in,

$g (x) = f (x) 2$

$1 + 3 x = f (x) 2$

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

$m a r i l y n = m a r i l y n - 1 + m a r i l y n - 2$

$\sqrt 2$

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

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

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