1. Problem statement: Compute the indefinite integrals in 4(a), evaluate the indefinite integrals in 4(b), solve the antiderivative and area problems in 5, and carry out the differentiation and related tasks in 6 as given. 2. 4(a)(i) Indefinite integral $\int(\sqrt{x}-\sin(3x/2)-e^{-2x})\,dx$. 2.1 Formula and rules: Use linearity of the integral and the power rule $\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ for $n\neq-1$ and $\int\sin(kx)\,dx=-\frac{1}{k}\cos(kx)$ and $\int e^{ax}\,dx=\frac{1}{a}e^{ax}$. 2.2 Work: Write $\sqrt{x}=x^{1/2}$ so $\int x^{1/2}\,dx=\frac{2}{3}x^{3/2}$. 2.3 Work: For the term $-\sin(3x/2)$ we have $\int -\sin(3x/2)\,dx= -\big(-\frac{2}{3}\cos(3x/2)\big)=\frac{2}{3}\cos\big(\tfrac{3x}{2}\big)$. 2.4 Work: For $-e^{-2x}$ use $\int e^{-2x}\,dx=-\frac{1}{2}e^{-2x}$ so $\int -e^{-2x}\,dx=\frac{1}{2}e^{-2x}$. 2.5 Combine results: $\displaystyle \int(\sqrt{x}-\sin(3x/2)-e^{-2x})\,dx=\frac{2}{3}x^{3/2}+\frac{2}{3}\cos\big(\tfrac{3x}{2}\big)+\frac{1}{2}e^{-2x}+C$. 3. 4(a)(ii) Indefinite integral $\int\big(\tfrac{5}{4}x-\sec^{2}x+3x^{2}\big)\,dx$. 3.1 Clarification and rule: Interpret $\tfrac{5}{4}x$ as $\frac{5}{4}x$ and use $\int\sec^{2}x\,dx=\tan x$. 3.2 Work: $\int\frac{5}{4}x\,dx=\frac{5}{8}x^{2}$. 3.3 Work: $\int -\sec^{2}x\,dx=-\tan x$. 3.4 Work: $\int 3x^{2}\,dx=x^{3}$. 3.5 Combine results: $\displaystyle \int\big(\tfrac{5}{4}x-\sec^{2}x+3x^{2}\big)\,dx=\frac{5}{8}x^{2}-\tan x+x^{3}+C$. 4. 4(b)(i) Definite integral $\displaystyle \int_{2}^{5}\big(\ln(e^{3x})+\tfrac{2}{x}-2\big)\,dx$. 4.1 Simplify integrand: $\ln(e^{3x})=3x$ so integrand becomes $3x+\tfrac{2}{x}-2$. 4.2 Antiderivative: $\int 3x\,dx=\tfrac{3}{2}x^{2}$, $\int \tfrac{2}{x}\,dx=2\ln|x|$, $\int -2\,dx=-2x$. 4.3 Evaluate: $F(x)=\tfrac{3}{2}x^{2}+2\ln|x|-2x$. 4.4 Compute $F(5)-F(2)$: $F(5)=\tfrac{75}{2}+2\ln 5-10=\tfrac{55}{2}+2\ln 5$ and $F(2)=6+2\ln 2-4=2+2\ln 2$. 4.5 Final value: $\displaystyle \int_{2}^{5}\big(\ln(e^{3x})+\tfrac{2}{x}-2\big)\,dx=\frac{55}{2}+2\ln 5-\big(2+2\ln 2\big)=\frac{51}{2}+2\ln\big(\tfrac{5}{2}\big)$. 5. 4(b)(ii) Definite integral $\displaystyle \int_{-\pi}^{\pi}\big(x^{4}-e^{-x}+\cos(3x)\big)\,dx$. 5.1 Use symmetry: $x^{4}$ is even so $\int_{-\pi}^{\pi}x^{4}\,dx=2\int_{0}^{\pi}x^{4}\,dx$. 5.2 $\cos(3x)$ is even and $\int_{-\pi}^{\pi}\cos(3x)\,dx=0$ because $\sin(3\pi)-\sin(0)=0$. 5.3 For $e^{-x}$ no parity cancellation; compute $\int_{-\pi}^{\pi}e^{-x}\,dx=\big[-e^{-x}\big]_{-\pi}^{\pi}=e^{\pi}-e^{-\pi}$. 5.4 Compute polynomial part: $\int_{0}^{\pi}x^{4}\,dx=\frac{\pi^{5}}{5}$ so even part gives $\frac{2\pi^{5}}{5}$. 5.5 Final value: $\displaystyle \int_{-\pi}^{\pi}\big(x^{4}-e^{-x}+\cos(3x)\big)\,dx=\frac{2\pi^{5}}{5}-\big(e^{\pi}-e^{-\pi}\big)$. 6. 5(a) Find $h(x)$ given $h'(x)=4x^{3}-x$ and $h(1)=5$. 6.1 Antiderivative: $\int\big(4x^{3}-x\big)\,dx=x^{4}-\tfrac{1}{2}x^{2}+C$. 6.2 Use condition $h(1)=1-\tfrac{1}{2}+C=\tfrac{1}{2}+C=5$ so $C=\tfrac{9}{2}$. 6.3 Final function: $\displaystyle h(x)=x^{4}-\tfrac{1}{2}x^{2}+\tfrac{9}{2}$. 7. 5(b)(i) Show algebraically intersections of $g(x)=-x^{2}-2x$ and $f(x)=-x^{3}$. 7.1 Set equal: $-x^{2}-2x=-x^{3}$ which gives $x^{3}-x^{2}-2x=0$. 7.2 Factor: $x(x^{2}-x-2)=x(x-2)(x+1)=0$. 7.3 Roots: $x=0,2,-1$. Hence the functions intersect at $x=-1,0,2$. 8. 5(b)(ii) Area of green region between the curves from $x=-1$ to $0$. 8.1 Determine upper function on $[-1,0]$ by sample or algebra; $g(x)=-x^{2}-2x$ and $f(x)=-x^{3}$ and $g(x)-f(x)=x^{3}-x^{2}-2x$ which is positive on $(-1,0)$ so $g$ is above $f$. 8.2 Area $A=\int_{-1}^{0}\big(g(x)-f(x)\big)\,dx=\int_{-1}^{0}\big(x^{3}-x^{2}-2x\big)\,dx$. 8.3 Antiderivative: $\int\big(x^{3}-x^{2}-2x\big)\,dx=\tfrac{1}{4}x^{4}-\tfrac{1}{3}x^{3}-x^{2}$. 8.4 Evaluate from $-1$ to $0$: at $0$ gives $0$ and at $-1$ gives $\tfrac{1}{4}-\big(-\tfrac{1}{3}\big)-1=\tfrac{1}{4}+\tfrac{1}{3}-1= -\tfrac{5}{12}$. 8.5 Area $A=0-(-\tfrac{5}{12})=\tfrac{5}{12}$ square units. 9. 5(c) If $\int_{1}^{4}f(x)\,dx=14$ and $\int_{1}^{3}f(x)\,dx=8$ then $\int_{3}^{4}f(x)\,dx=14-8=6$. 10. 5(d) Find $y$ so that $\displaystyle \int_{-\sqrt{y}}^{\sqrt{y}}(x+4)\,dx=8\sqrt{3}$. 10.1 Use linearity: odd part $\int_{-a}^{a}x\,dx=0$ and even part $\int_{-a}^{a}4\,dx=8a$ with $a=\sqrt{y}$. 10.2 So integral equals $8\sqrt{y}$ and set $8\sqrt{y}=8\sqrt{3}$ hence $\sqrt{y}=\sqrt{3}$ and $y=3$. 11. 6(a)(i) Differentiate $f(x)=4x^{2}-9x$ from first principles. 11.1 Definition: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. 11.2 Compute $f(x+h)=4(x+h)^{2}-9(x+h)=4x^{2}+8xh+4h^{2}-9x-9h$. 11.3 Subtract $f(x)$ and divide by $h$ gives $\frac{h(8x+4h-9)}{h}=8x+4h-9$. 11.4 Take limit $h\to0$ to get $f'(x)=8x-9$. 12. 6(a)(ii) Differentiate $g(x)=\dfrac{1}{x-2}$ from first principles. 12.1 Compute difference quotient: $\frac{1}{x+h-2}-\frac{1}{x-2}=\frac{-h}{(x+h-2)(x-2)}$. 12.2 Divide by $h$ gives $-\frac{1}{(x+h-2)(x-2)}$ and take $h\to0$ to get $g'(x)=-\frac{1}{(x-2)^{2}}$. 13. 6(b) For $y=\ln(-x)$ show $\dfrac{dy}{dx}=\big(\dfrac{dx}{dy}\big)^{-1}$. 13.1 Solve for $x$ in terms of $y$: $y=\ln(-x)\Rightarrow -x=e^{y}\Rightarrow x=-e^{y}$. 13.2 Compute $\dfrac{dx}{dy}=-e^{y}$ so $\big(\dfrac{dx}{dy}\big)^{-1}=-e^{-y}$. 13.3 Compute $\dfrac{dy}{dx}=\dfrac{d}{dx}\ln(-x)=\frac{1}{-x}\cdot(-1)=\frac{1}{x}$ and using $x=-e^{y}$ we get $\dfrac{1}{x}=-e^{-y}$. 13.4 Hence $\dfrac{dy}{dx}=\big(\dfrac{dx}{dy}\big)^{-1}$ as required. 14. 6(c) Normal to $y=x^{4}-5x^{2}+4$ at $x=-1$. 14.1 Derivative $y'=4x^{3}-10x$. 14.2 Slope of tangent at $x=-1$ is $y'(-1)=4(-1)^{3}-10(-1)=-4+10=6$. 14.3 Slope of normal is $-\tfrac{1}{6}$ and point is $(-1,0)$ since $y(-1)=1-5+4=0$. 14.4 Equation of normal: $y-0=-\tfrac{1}{6}(x+1)$ or $\displaystyle y=-\tfrac{1}{6}x-\tfrac{1}{6}$. 15. 6(d) Show $\displaystyle \int 6x\cos(x^{2})\,dx=3\sin(x^{2})+k$. 15.1 Use substitution $u=x^{2}$ so $du=2x\,dx$ and $6x\,dx=3\,du$. 15.2 Then $\int 6x\cos(x^{2})\,dx=\int 3\cos(u)\,du=3\sin(u)+C=3\sin(x^{2})+k$. Final answers summary: - 4(a)(i) $\displaystyle \frac{2}{3}x^{3/2}+\frac{2}{3}\cos\big(\tfrac{3x}{2}\big)+\frac{1}{2}e^{-2x}+C$. - 4(a)(ii) $\displaystyle \frac{5}{8}x^{2}-\tan x+x^{3}+C$. - 4(b)(i) $\displaystyle \frac{51}{2}+2\ln\big(\tfrac{5}{2}\big)$. - 4(b)(ii) $\displaystyle \frac{2\pi^{5}}{5}-\big(e^{\pi}-e^{-\pi}\big)$. - 5(a) $\displaystyle h(x)=x^{4}-\tfrac{1}{2}x^{2}+\tfrac{9}{2}$. - 5(b)(i) Intersections at $x=-1,0,2$. - 5(b)(ii) Area $\displaystyle \frac{5}{12}$. - 5(c) $\displaystyle 6$. - 5(d) $\displaystyle y=3$. - 6(a)(i) $\displaystyle f'(x)=8x-9$. - 6(a)(ii) $\displaystyle g'(x)=-\frac{1}{(x-2)^{2}}$. - 6(b) Verified $\displaystyle \frac{dy}{dx}=\Big(\frac{dx}{dy}\Big)^{-1}$. - 6(c) Normal $\displaystyle y=-\tfrac{1}{6}x-\tfrac{1}{6}$. - 6(d) $\displaystyle 3\sin(x^{2})+k$.