QCAA Specialist Mathematics Alternative Sequence Applications of integral calculus

Finding the points of intersection of the two functions
$y = 4x$ and $y = 2x^2$
$4x = 2x^2$
$2x^2 - 4x = 0 \Rightarrow 2x(x - 2) = 0 \Rightarrow x = 0$ and $x = 2$
When $x = 0, y = 0$
When $x = 2, y = 8$
Rearranging the two functions in the form $x = f(y)$ and $x = g(y)$
$x = \frac{y}{4}$ and $x = \pm\sqrt{\frac{y}{2}}$
Finding volume of revolution between curves
$V = \left| \pi \int_a^b [f(y)]^2 - [g(y)]^2 \, dy \right|$
$= \left| \pi \int_0^8 \frac{y}{2} - \frac{y^2}{16} \, dy \right|$
$= \pi \left| \frac{y^2}{4} - \frac{y^3}{48} \right|_0^8$
$= \pi \left| \left(16 - \frac{32}{3}\right) - (0) \right|$
$= \frac{16\pi}{3}$

Given $x = \tan^{-1}(t - 1)$
$t = \tan(x) + 1$

Substituting into $y = t - 2 + 2t^{-1}$
$y = \tan(x) + 1 - 2 + \frac{2}{\tan(x) + 1}$

$= \frac{(\tan(x) - 1)(\tan(x) + 1) + 2}{\tan(x) + 1}$

$= \frac{\tan^2(x) + 1}{\tan(x) + 1}$

$= \frac{\sec^2(x)}{\tan(x) + 1}$

Area $= \int_a^b \frac{\sec^2(x)}{\tan(x) + 1} \ dx$

Let $u = \tan(x) + 1 \Rightarrow \frac{du}{dx} = \sec^2(x)$

Area $= \int_{x=a}^{x=b} \frac{du}{u} = \ln|u| \ \Big|_{x=a}^{x=b} = \ln(\tan(x) + 1) \ \Big|_a^b$

$= [\ln(\tan(b) + 1) - \ln(\tan(a) + 1)]$

$= \ln \left( \frac{\tan(b) + 1}{\tan(a) + 1} \right)$

Given $a = 0$ and $b = 1$
Area $= \ln \left( \frac{\tan(1) + 1}{\tan(0) + 1} \right) \approx 0.94 \text{ units}^2$

Using GDC
Area $= \int_0^1 \frac{\sec^2(x)}{\tan(x) + 1} \ dx \approx 0.94 \text{ units}^2$

The result is verified for this example.