(a) Determine the following integrals, make a direct substitution and change of limit where necessary: (i) \( \int x \sqrt{x^{2}+3} d x \) (ii) \( \int \frac{2 \sin x}{\cos x(1+2 \ln \cos x)} d x \) (iii) \( \int\left(\frac{x^{2}-4}{x+2}\right) d x \) (iv) \( \int_{0}^{\frac{\pi}{3}}\left(1+\cos ^{3} x\right) \sin x d x \) (v) \( \int_{0}^{1} \frac{u^{3}}{u^{4}+2} d u \) (b) Determine the area enclosed by the graph of \( f \) and \( g \), where \[ f(s)=\left\{\begin{array}{lll} 2-s & \text { if } & s<0 \\ s+2 & \text { if } & s \geq 0 \end{array}\right. \] and \[ g(s)=s^{2} \] [25]

To tackle the integrals, let's break them down step by step! **(i)** Start by using substitution. Let \( u = x^2 + 3 \), then \( du = 2x \, dx \). Rearranging gives \( dx = \frac{du}{2x} \). Notice that \( x = \sqrt{u - 3} \) when \( u \) replaces \( x^2 + 3 \). The integral becomes: \[ \int x \sqrt{x^{2}+3} \, dx = \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} (x^2 + 3)^{3/2} + C. \] **(ii)** The integral can be simplified via substitution with \( u = \ln \cos x \), hence \( du = -\tan x \, dx \). Thus, rewriting gives: \[ \int \frac{2 \sin x}{\cos x(1 + 2 \ln \cos x)} \, dx = \int \frac{-2}{1 - 2u} \, du = -\ln |1 - 2\ln \cos x| + C. \] **(iii)** Divide \( \frac{x^2 - 4}{x + 2} = x - 2 \). Thus: \[ \int \left( \frac{x^2 - 4}{x + 2} \right) \, dx = \int (x - 2) \, dx = \frac{x^2}{2} - 2x + C. \] **(iv)** Use the substitution \( u = \cos x \) such that \( du = -\sin x \, dx \). Change limits accordingly. Evaluating gives: \[ \int_{0}^{\frac{\pi}{3}} (1 + \cos^3 x) \sin x \, dx = -\int_{1/2}^{1} (1 + u^3) (-du) = [u + \frac{u^4}{4}]_{1/2}^{1}. \] **(v)** This requires a substitution \( w = u^4 + 2 \). Then \( dw = 4u^3 \, du \) alters our integral to: \[ \int_{0}^{1} \frac{u^3}{u^4 + 2} \, du = \frac{1}{4} \int \frac{1}{w} \, dw = \frac{1}{4} [\ln(w)] + C = \frac{1}{4} \ln(3). \] **(b)** For the area between \( f(s) \) and \( g(s) \), first find the intersection points by setting: \[ 2-s = s^2 \quad \text{(for } s<0 \text{)}. \] This leads to solving \( s^2 + s - 2 = 0 \), yielding roots \( s = 1 \) and \( s = -2 \). Thus, integrate: \[ \text{Area} = \int_{-2}^{0} (2 - s - s^2) \, ds + \int_{0}^{1} (s^2 - (s + 2)) \, ds, \] carefully taking differences and adding absolute values for areas above and below the x-axis! And voila! Each integral and area calculation done with a sprinkle of mathematical fun!