5 (a) Solve the following integral.

1. Express the numerator in terms of the derivative of the quadratic expression in the denominator to split the integral.

$\frac{\mathrm{d}}{\mathrm{d}x}(3+2x-x^2) = 2-2x$
$\int \frac{1-2x}{\sqrt{3+2x-x^2}}\,\mathrm{d}x = \int \frac{(2-2x) - 1}{\sqrt{3+2x-x^2}}\,\mathrm{d}x$
$= \int \frac{2-2x}{\sqrt{3+2x-x^2}}\,\mathrm{d}x - \int \frac{1}{\sqrt{3+2x-x^2}}\,\mathrm{d}x$

2. Complete the square for the quadratic expression in the second integral.

$3+2x-x^2 = - (x^2 - 2x) + 3$
$= - \left[ (x-1)^2 - 1 \right] + 3$
$= 4 - (x-1)^2$
$\int \frac{1}{\sqrt{3+2x-x^2}}\,\mathrm{d}x = \int \frac{1}{\sqrt{2^2 - (x-1)^2}}\,\mathrm{d}x$

3. Evaluate both integrals using standard forms and combine the results.

For the first integral, using $\int \mathrm{f}'(x)[\mathrm{f}(x)]^n\,\mathrm{d}x$:
$\int (2-2x)(3+2x-x^2)^{-\frac{1}{2}}\,\mathrm{d}x = \frac{(3+2x-x^2)^{\frac{1}{2}}}{\frac{1}{2}} = 2\sqrt{3+2x-x^2}$

For the second integral, using $\int \frac{1}{\sqrt{a^2-x^2}}\,\mathrm{d}x = \sin^{-1}\left(\frac{x}{a}\right)$:
$\int \frac{1}{\sqrt{2^2 - (x-1)^2}}\,\mathrm{d}x = \sin^{-1}\left(\frac{x-1}{2}\right)$

Combining the results and adding the constant of integration:
$= 2\sqrt{3+2x-x^2} - \sin^{-1}\left(\frac{x-1}{2}\right) + C$

1. Apply integration by parts to evaluate the integral.

Let $u = \ln(2-x^2)$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = x$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{-2x}{2-x^2}$ and $v = \frac{1}{2}x^2$.

Using the integration by parts formula $\int u \frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x$:
$\int x\ln\left(2-x^2\right)\,\mathrm{d}x = \frac{1}{2}x^2\ln\left(2-x^2\right) - \int \frac{1}{2}x^2\left(\frac{-2x}{2-x^2}\right)\,\mathrm{d}x$
$= \frac{1}{2}x^2\ln\left(2-x^2\right) + \int \frac{x^3}{2-x^2}\,\mathrm{d}x$

2. Use algebraic manipulation to rewrite the remaining improper rational expression.

Rewrite the improper fraction $\frac{x^3}{2-x^2}$ by performing polynomial division or splitting the numerator:
$\frac{x^3}{2-x^2} = \frac{-x(2-x^2) + 2x}{2-x^2} = -x + \frac{2x}{2-x^2}$

Therefore, the integral becomes:
$\int \frac{x^3}{2-x^2}\,\mathrm{d}x = \int \left( -x + \frac{2x}{2-x^2} \right)\,\mathrm{d}x$

3. Integrate the simplified terms using standard forms to obtain the final result.

Using the standard form $\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)}\,\mathrm{d}x = \ln|\mathrm{f}(x)| + c$:
$\int \left( -x + \frac{2x}{2-x^2} \right)\,\mathrm{d}x = -\frac{1}{2}x^2 - \ln(2-x^2)$
(Note: The absolute value is not needed since the given domain is $-\sqrt{2}<x<\sqrt{2}$, which ensures $2-x^2 > 0$.)

Putting it all together and adding the constant of integration $C$:
$\int x\ln\left(2-x^2\right)\,\mathrm{d}x = \frac{1}{2}x^2\ln\left(2-x^2\right) - \frac{1}{2}x^2 - \ln(2-x^2) + C$
$= \frac{1}{2}(x^2-2)\ln(2-x^2) - \frac{1}{2}x^2 + C$

1. Determine the $x$- and $y$-intercepts of the graph $y = |f(x)|$.

For the $x$-intercept, set $y = 0$:
$|e^{-x} - 2| = 0$
$e^{-x} = 2$
$-x = \ln 2 \implies x = -\ln 2$

For the $y$-intercept, set $x = 0$:
$y = |e^0 - 2| = |1 - 2| = |-1| = 1$

2. Determine the equation of the horizontal asymptote.

Consider the behaviour of $f(x)$ as $x \to \infty$:
As $x \to \infty$, $e^{-x} \to 0$, so $e^{-x} - 2 \to -2$.
Therefore, as $x \to \infty$, $|f(x)| \to |-2| = 2$.
The horizontal asymptote is $y = 2$.

3. Describe the final shape of the modulus curve to be sketched.

The original graph $y = e^{-x} - 2$ is a strictly decreasing curve from $\infty$ to its horizontal asymptote at $y = -2$.
Applying the modulus transformation $y = |f(x)|$ reflects the portion of the curve where $x > -\ln 2$ (which is below the $x$-axis) in the $x$-axis.
The resulting sketch should decrease from the second quadrant to touch the $x$-axis at $(-\ln 2, 0)$, then turn to increase, crossing the $y$-axis at $(0, 1)$, and gently flatten out to approach the asymptote $y = 2$ from below.

1. Determine the expression for $|f(x)|$ over the interval $[0, \ln 3]$ using the behaviour of the graph.

From the graph in (b)(i), we know that for $x \ge 0$, the curve $y = e^{-x} - 2$ is entirely below the $x$-axis.
Thus, for the interval $0 \le x \le \ln 3$, $f(x) < 0$, which means:
$|f(x)| = -(e^{-x} - 2) = 2 - e^{-x}$

2. Set up and perform the definite integration.

$\int_{0}^{\ln 3} |f(x)| \,\mathrm{d}x = \int_{0}^{\ln 3} (2 - e^{-x}) \,\mathrm{d}x$
$= \left[ 2x + e^{-x} \right]_{0}^{\ln 3}$

3. Substitute the limits to calculate the exact value.

$= \left( 2\ln 3 + e^{-\ln 3} \right) - \left( 2(0) + e^{0} \right)$
$= \left( 2\ln 3 + \frac{1}{3} \right) - (0 + 1)$
$= 2\ln 3 - \frac{2}{3}$