QCAA Mathematical Methods Introduction to integration

$F(x) = \int \left(4e^{2x} + \sin(2x)\right) dx$
$= \frac{4e^{2x}}{2} + -\frac{1}{2}\cos(2x) + c$
$= 2e^{2x} - \frac{1}{2}\cos(2x) + c$
$F(0) = 5$
$5 = 2 - \frac{1}{2} + c$
$c = 3\frac{1}{2}$
$F(x) = 2e^{2x} - \frac{1}{2}\cos(2x) + 3\frac{1}{2}$

$\frac{dy}{dx} = \left(\frac{3x^7 - 2x}{x^4}\right)^2$
$= \left(\frac{3x^7}{x^4} - \frac{2x}{x^4}\right)^2$
$= \left(3x^3 - \frac{2}{x^3}\right)^2$
$= (3x^3)^2 - 2(3x^3)\left(\frac{2}{x^3}\right) + \left(\frac{2}{x^3}\right)^2$
$= 9x^6 - 12 + \frac{4}{x^6}$
$= 9x^6 - 12 + 4x^{-6}$
$y = \int (9x^6 - 12 + 4x^{-6}) dx$
$y = \frac{9}{7}x^7 - 12x - \frac{4}{5}x^{-5} + c$

$V = \int 0.25e^{0.25t} dt$
$= e^{0.25t} + c$
when $t = 0, V = 0$
$\therefore 0 = e^{0.25 \times 0} + c$
$\therefore c = -1$
$\therefore V = e^{0.25t} - 1$

$V(8\ln(6)) = e^{0.25 \times 8 \ln(6)} - 1$
$= 36 - 1$

$= 35$ litres

Using trapezoidal rule
Volume after 3 hours $= \frac{1}{2}(0.25+0.53+2(0.32+0.41))$

Volume after 3 hours $= 1.12$ litres

7 times the members is 7000.
Let m be the time when 7000 members is reached.
The required change in members is 6000.
$6000 = \int_0^m 3e^{0.5t} dt$
$= \left[ 3 \times \frac{1}{0.5}e^{0.5t} \right]_0^m$
$= \left[ 6e^{0.5t} \right]_0^m$
$= \left[ 6e^{0.5m} - 6e^0 \right]$
$= \left[ 6e^{0.5m} - 6 \right]$
$6000 = 6e^{0.5m} - 6$
$6006 = 6e^{0.5m}$
$\frac{6006}{6} = e^{0.5m}$
$0.5m = \ln 1001$
$m = 2\ln 1001 \text{ days}$