Use the substitution $u = ext{log}_e x$ to evaluate \[ \int_{e}^{e^2} \frac{1}{x \left( \text{log}_e x \right)^2} \, dx - HSC - SSCE Mathematics Extension 1 - Question 2 - 2008 - Paper 1

From the acceleration equation, we have: [ \bar{x} = \frac{d}{dt} \left( \frac{1}{2} v^2 \right) \Rightarrow \bar{x} = v \frac{dv}{dx}. ]

Setting the two equal yields: [ v \frac{dv}{dx} = 4 \Rightarrow v dv = 4 dx. ]

Integrating both sides: [ \int v , dv = \int 4 , dx \Rightarrow \frac{1}{2} v^2 = 4x + C. ]

Since the particle starts at rest ($v(1) = 0$), we find $C = -4$.

Thus, at $x = 2$: [ \frac{1}{2} v^2 = 4(2) - 4 \Rightarrow \frac{1}{2} v^2 = 4 \Rightarrow v^2 = 8 \Rightarrow v = \sqrt{8} = 2\sqrt{2}. ]

Using Vieta's formulas, the sum of the roots is given by: [ -\frac{16}{a} = -2 + 3 + \alpha \Rightarrow \alpha = -2 + 3 + \frac{16}{a}. ]

The product of the roots is: [ -\frac{-120}{a} = (-2)(3)(\alpha) \Rightarrow 6\alpha = \frac{120}{a} \Rightarrow \alpha = \frac{20}{a}. ]

Setting these two expressions for ( \alpha ) equal gives: [ -2 + 3 + \frac{16}{a} = \frac{20}{a}. ]

Simplifying, we find: [ 1 + \frac{16}{a} = \frac{20}{a} \Rightarrow 4 = \frac{4}{a} \Rightarrow a = 4. ] [ \alpha = \frac{20}{4} = 5. ]

First, find the derivative of $f(x)$: [ f'(x) = \sec^2(x) - \frac{1}{x}. ]

Starting with $x_0 = 4$, compute: [ f(4) = \tan(4) - \log_e(4), ] [ f'(4) = \sec^2(4) - \frac{1}{4}. ]

Use Newton's method to find: [ x_{1} = x_{0} - \frac{f(x_0)}{f'(x_0)}. ]

Continue this process until the value stabilizes, rounding your final answer to two decimal places.