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:

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

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

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

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

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

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

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

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

$$g(t) = 1 + 3t$$

Plugging back in,

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

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

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

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

$$\sqrt 2$$

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

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

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