(i)

\begin{align} \text{Area of shaded region } A & = \int_1^p {8 \over x^2} \phantom{0} dx \\ & = \int_1^p 8x^{-2} \phantom{0} dx \\ & = \left[ {8x^{-1} \over -1} \right]_1^p \\ & = \left[ -8x^{-1} \right]_1^p \\ & = \left[ -{8 \over x} \right]_1^p \\ & = -{8 \over p} - \left(-{8 \over 1} \right) \\ & = -{8 \over p} + 8 \\ & = \left( 8 - {8 \over p} \right) \text{ units}^2 \end{align}

(ii)

\begin{align} \text{Area of shaded region } B & = \int_p^6 {8 \over x^2} \phantom{0} dx \\ & = \left[ -{8 \over x} \right]_p^6 \\ & = -{8 \over 6} - \left(-{8 \over p} \right) \\ & = -{4 \over 3} + {8 \over p} \\ & = \left( {8 \over p} - {4 \over 3} \right) \text{ units}^2 \\ \\ \text{Area of region } A & = \text{Area of region } B \\ 8 - {8 \over p} & = {8 \over p} - {4 \over 3} \\ -{8 \over p} - {8 \over p} & = -{4 \over 3} - 8 \\ -{16 \over p} & = -{28 \over 3} \\ {16 \over p} & = {28 \over 3} \\ 16(3) & = 28p \\ 48 & = 28p \\ {48 \over 28} & = p \\ {12 \over 7} & = p \end{align}

(a)

\begin{align} u & = x &&& v & = \sin x \\ {du \over dx} & = 1 &&& {dv \over dx} & = \cos x \end{align} \begin{align} {d \over dx} (x \sin x) & = (x)(\cos x) + (\sin x)(1) \phantom{000000} [\text{Product rule}] \\ & = x \cos x + \sin x\\ \\ \therefore {d \over dx}(x \sin x + \cos x) & = (x \cos x + \sin x) + (- \sin x) \\ & = x \cos x + \sin x - \sin x \\ & = x \cos x \phantom{00} \text{ (Shown)} \end{align}

(b)(i) Point $A$ is the $x$-intercept of the curve.

\begin{align} y & = x \cos x \\ \\ \text{Let } & y = 0, \\ 0 & = x \cos x \\ 0 & = (x)(\cos x) \\ \\ x = 0 \phantom{00} & \text{or} \phantom{000} \cos x = 0 \\ \\ \\ \cos x & = 0 \end{align}

\begin{align} x & = {\pi \over 2} , \phantom{0} {3\pi \over 2} \\ \\ \therefore & \phantom{.} A \left( {\pi \over 2}, 0 \right) \end{align}

(b)(ii)

\begin{align} \text{From (a), } {d \over dx}(x\sin x + \cos x) & = x \cos x \\ \\ \implies \int x \cos x \phantom{0} dx & = x\sin x + \cos x \\ \\ \text{Area of shaded region} & = \int_0^{\pi \over 2} x\cos x \phantom{0} dx \\ & = \left[ x\sin x + \cos x \right]_0^{\pi \over 2} \\ & = \left[ {\pi \over 2}\left(\sin {\pi \over 2}\right) + \cos {\pi \over 2} \right] - \left[ (0)(\sin 0) + \cos 0 \right] \\ & = \left[ {\pi \over 2} (1) + (0) \right] - \left[ 0 + 1 \right] \\ & = {\pi \over 2} - 1 \\ & = 0.57079 \\ & \approx 0.571 \text{ units}^2 \end{align}

\begin{align} y^2 & = 3x \\ \\ \text{When } & x = 3, \\ y^2 & = 3(3) \\ & = 9 \\ y & = \pm \sqrt{9} \\ & = \pm 3 \\ \\ y \text{-coordinates of } & \text{point of intersection are } 3 \text{ and } - 3 \\ \\ \text{Area bounded by line (w.r.t }y \text{-axis}) & = \int_{-3}^3 3 \phantom{.} dy \\ & = [3y]_{-3}^3 \\ & = [3(3)] - [3(-3)] \\ & = 9 - (-9) \\ & = 18 \text{ units}^2 \\ \\ \text{Eqn of curve: } & y^2 = 3x \\ {1 \over 3}y^2 & = x \phantom{000000} [\text{Make } x \text{ the subject}] \\ \\ \text{Area bounded by curve (w.r.t } y \text{-axis}) & = \int_{-3}^3 {1 \over 3}y^2 \phantom{.} dy \\ & = \left[ {1 \over 3} \left(y^3 \over 3\right) \right]_{-3}^3 \\ & = \left[ {y^3 \over 9} \right]_{-3}^3 \\ & = \left[ (3)^3 \over 9 \right] - \left[ (-3)^3 \over 9\right] \\ & = 3 - (-3) \\ & = 6 \text{ units}^2 \\ \\ \text{Area of shaded region} & = 18 - 6 \\ & = 12 \text{ units}^2 \end{align}

Note region $B$ is below the $x$-axis

\begin{align} \text{Area of shaded region } A & = \int_0^1 x^3 - 3x^2 + 2x \phantom{0} dx \\ & = \left[ {x^4 \over 4} - {3x^3 \over 3} + {2x^2 \over 2} \right]_0^1 \\ & = \left[ {x^4 \over 4} - x^3 + x^2 \right]_0^1 \\ & = \left[ {(1)^4 \over 4} - (1)^3 + (1)^2 \right] - \left[ {(0)^4 \over 4} - (0)^3 + (0)^2 \right] \\ & = {1 \over 4} - 0 \\ & = {1 \over 4} \text{ units}^2 \\ \\ \text{Area of shaded region } B & = - \int_1^2 (x^3 - 3x^2 + 2x) \phantom{0} dx \\ & = - \left[ {x^4 \over 4} - x^3 + x^2 \right]_1^2 \\ & = - \left\{ \left[ {(2)^4 \over 4} - (2)^3 + (2)^2 \right] - \left[ {(1)^4 \over 4} - (1)^3 + (1)^2 \right] \right\} \\ & = - \left[ 0 - {1 \over 4} \right] \\ & = {1 \over 4} \text{ units}^2 \\ & = \text{Area of shaded region } A \phantom{00} \text{ (Shown)} \end{align}

(i) Point $P$ is the $y$-intercept of the curve while the $x$-coordinate of $Q$ is $3$.

\begin{align} y & = x^2 - 2x + 2 \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 - 2(0) + 2 \\ & = 2 \\ \\ \therefore & \phantom{.} P(0, 2) \\ \\ \\ \text{Let } & x = 3, \\ y & = (3)^2 - 2(3) + 2 \\ & = 5 \\ \\ \therefore & \phantom{.} Q(3, 5) \end{align}

(ii) The 'sum' of regions A and B is a trapezium

\begin{align} \text{Area of shaded region } A & = \int_0^3 x^2 - 2x + 2 \phantom{0} dx \\ & = \left[ {x^3 \over 3} - {2x^2 \over 2} + 2x \right]_0^3 \\ & = \left[ {x^3 \over 3} - x^2 + 2x \right]_0^3 \\ & = \left[ {(3)^3 \over 3} - (3)^2 + 2(3) \right] - \left[ {(0)^3 \over 3} - (0)^2 + 2(0) \right] \\ & = 6 - 0 \\ & = 6 \text{ units}^2 \\ \\ \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides } \times \text{Height} \\ & = {1 \over 2} \times (2 + 5) \times (3) \\ & = 10{1 \over 2} \text{ units}^2 \\ \\ \therefore \text{Area of shaded region } B & = 10{1 \over 2} - 6 \\ & = 4{1 \over 2} \text{ units}^2 \end{align}

(i) Point $A$ is the point of intersection between the curve and the line while point $B$ is the $x$-intercept of the curve.

\begin{align} y & = 2x \phantom{0} \text{ --- (1)} \\ \\ y & = 6x - x^2 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x & = 6x - x^2 \\ x^2 - 4x & = 0 \\ x(x - 4) & = 0 \\ \\ x = 0 \phantom{00} & \text{or} \phantom{000} x - 4 = 0 \\ & \phantom{or000-4} x = 4 \\ \\ x-\text{cooordinate of } A & = 4 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = 2(4) \\ & = 8 \\ \\ \therefore & \phantom{.} A(4, 8) \\ \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = 6x - x^2 \\ 0 & = x(6 - x) \\ \\ x = 0 \phantom{00} & \text{or} \phantom{000} 6 - x = 0 \\ & \phantom{or0006} - x = -6 \\ & \phantom{or0006(0)} x = 6 \\ \\ \therefore & \phantom{.} B (6, 0) \end{align}

(ii)

Region $C$ is the area bounded by the line from $x = 0$ to $x = 4$ and is a right-angled triangle

Region $D$ is the area bounded by the curve from $x = 4$ to $x = 6$

\begin{align} \text{Area of region C} & = {1 \over 2} \times \text{Base } \times \text{Height} \\ & = {1 \over 2} \times 4 \times 8 \\ & = 16 \text{ units}^2 \\ \\ \text{Area of region D} & = \int_4^6 (6x - x^2) \phantom{0} dx \\ & = \left[ {6x^2 \over 2} - {x^3 \over 3} \right]_4^6 \\ & = \left[ 3x^2 - {x^3 \over 3} \right]_4^6 \\ & = \left[ 3(6)^2 - {(6)^3 \over 3} \right] - \left[ 3(4)^2 - {(4)^3 \over 3} \right] \\ & = 36 - 26{2 \over 3} \\ & = 9{1 \over 3} \text{ units}^2 \\ \\ \therefore \text{Area of shaded region} & = 16 + 9{1 \over 3} \\ & = 25{1 \over 3} \text{ units}^2 \end{align}

(i)

\begin{align} y & = x^2 - 4x + 5 \\ \\ \text{When } & x = 3, \\ y & = (3)^2 - 4(3) + 5 \\ & = 2 \\ \\ \therefore & \phantom{.} P(3, 2) \\ \\ \\ y & = x^2 - 4x + 5 \\ \\ {dy \over dx} & = 2x - 4 \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = 2(3) - 4 \\ & = 6 - 4 \\ & = 2 \\ \\ \text{Gradient of normal} & = {-1 \over 2} \phantom{000000} [\text{Perpendicular lines}] \\ & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } & P(3, 2), \\ 2 & = -{1 \over 2}(3) + c \\ 2 & = -{3 \over 2} + c \\ {7 \over 2} & = c \\ \\ y & = -{1 \over 2}x + {7 \over 2} \\ 2y & = -x + 7 \\ x + 2y & = 7 \phantom{00} \text{ (Shown)} \end{align}

(ii) Point $Q$ is the $x$-intercept of the normal.

\begin{align} \text{Eqn of normal: } & y = -{1 \over 2}x + {7 \over 2} \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 2}x + {7 \over 2} \\ {1 \over 2}x & = {7 \over 2} \\ x & = 7 \\ \\ \therefore & \phantom{.} Q(7, 0) \end{align}

(iii)

\begin{align} \text{Area of region A } & = \int_0^3 (x^2 - 4x + 5) \phantom{0} dx \\ & = \left[ {x^3 \over 3} - {4x^2 \over 2} + 5x \right]_0^3 \\ & = \left[ {x^3 \over 3} - 2x^2 + 5x \right]_0^3 \\ & = \left[ {(3)^3 \over 3} - 2(3)^2 + 5(3) \right] - \left[ {(0)^3 \over 3} - 2(0)^2 + 5(0) \right] \\ & = 6 - 0 \\ & = 6 \text{ units}^2 \\ \\ \text{Area of region B } & = {1 \over 2} \times \text{Base } \times \text{Height} \\ & = {1 \over 2} \times 4 \times 2 \\ & = 4 \text{ units}^2 \\ \\ \therefore \text{Area of shaded region } & = 6 + 4 \\ & = 10 \text{ units}^2 \end{align}

(i)

\begin{align} y & = x(5 - x) \\ & = 5x - x^2 \\ & = -x^2 + 5x \phantom{000000} [\cap \text{ shape}] \\ \\ \text{Let } & y = 0, \\ 0 & = x(5 - x) \\ \\ x = 0 \phantom{00} & \text{or} \phantom{000} 5 - x = 0 \\ & \phantom{or0005} - x = -5 \\ & \phantom{or00050()} x = 5 \\ \\ \implies & x \text{-intercepts are } (0, 0) \text{ and } (5, 0) \\ \\ \text{Line of symmetry, } x & = {0 + 5 \over 2} \\ x & = 2.5 \\ \\ \text{Substitute } & x = 2.5 \text{ into eqn of curve,} \\ y & = (2.5)(5 - 2.5) \\ & = 6.25 \\ \\ \implies & \text{Maximum point is } (2.5, 6.25) \end{align}

\begin{align} \text{Area of left region} & = \int_0^2 -x^2 + 5x \phantom{0} dx \\ & = \left[ -{x^3 \over 3} + {5x^2 \over 2} \right]_0^2 \\ & = \left[ -{(2)^3 \over 3} + {5(2)^2 \over 2} \right] - \left[ -{(0)^3 \over 3} + {5(0)^2 \over 2} \right] \\ & = 7{1 \over 3} - 0 \\ & = {22 \over 3} \text{ units}^2 \\ \\ \text{Area of right region} & = \int_2^5 (-x^2 + 5x) \phantom{0} dx \\ & = \left[ -{x^3 \over 3} + {5x^2 \over 2} \right]_2^5 \\ & = \left[ -{(5)^3 \over 3} + {5(5)^2 \over 2} \right] - \left[ -{(2)^3 \over 3} + {5(2)^2 \over 2} \right] \\ & = 20{5 \over 6} - 7{1 \over 3} \\ & = 13{1 \over 2} \\ & = {27 \over 2} \text{ units}^2 \\ \\ {22 \over 3} & : {27 \over 2} \\ 6 \times {22 \over 3} & : {27 \over 2} \times 6 \\ 44 & : 81 \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} y & = x(x - 3) \\ & = x^2 - 3x \phantom{000000} [\cup \text{ shape}] \\ \\ \text{Let } & y = 0, \\ 0 & = x(x - 3) \\ \\ x = 0 \phantom{00} & \text{or} \phantom{000} x - 3 = 0 \\ & \phantom{or000-3} x = 3 \\ \\ \implies & x \text{-intercepts are } (0, 0) \text{ and } (3, 0) \\ \\ \text{Line of symmetry, } x & = {0 + 3 \over 2} \\ x & = 1.5 \\ \\ \text{Substitute } & x = 1.5 \text{ into eqn of curve,} \\ y & = (1.5)(1.5 - 3) \\ & = -2.25 \\ \\ \implies & \text{Minimum point is } (1.5, -2.25) \end{align}

\begin{align} \text{Substitute } & y = 2 \text{ into eqn of curve,} \\ -2 & = x^2 - 3x \\ 0 & = x^2 - 3x + 2 \\ 0 & = (x - 1)(x - 2) \\ \\ x - 1 = 0 \phantom{00} & \text{or} \phantom{000} x - 2 = 0 \\ x = 1 \phantom{00} & \phantom{or000-2} x = 2 \\ \\ \text{Line } y = 2 & \text{ meets curve at } (1, -2) \text{ and } (2, -2) \\ \\ \text{Area bounded by curve} & = - \int_1^2 x^2 - 3x \phantom{.} dx \\ & = - \left[ {x^3 \over 3} - {3x^2 \over 2} \right]_1^2 \\ & = - \left\{ \left[ {(2)^3 \over 3} - {3(2)^2 \over 2} \right] - \left[ {(1)^3 \over 3} - {3(1)^2 \over 2} \right] \right\} \\ & = - \left[ -3{1 \over 3} - \left(-1{1 \over 6}\right) \right] \\ & = 2{1 \over 6} \text{ units}^2 \\ \\ \text{Area of rectangle} & = 1 \times 2 \\ & = 2 \text{ units}^2 \\ \\ \text{Area of region bounded} & = 2{1 \over 6} - 2 \\ & = {1 \over 6} \text{ units}^2 \end{align}

(i)

\begin{align} u & = x - 1 &&& v & = e^x \\ {du \over dx} & = 1 &&& {dv \over dx} & = e^x \end{align} \begin{align} {d \over dx}[(x - 1)e^x] & = (x - 1)(e^x) + (e^x)(1) \phantom{000000} [\text{Product rule}] \\ & = e^x (x - 1) + e^x \\ & = x e^x - e^x + e^x \\ & = x e^x \phantom{00} \text{ (Shown)} \end{align}

(ii)

Regions B and C form a rectangle

\begin{align} {d \over dx}[(x - 1)e^x] & = x e^x \\ \\ \implies \int x e^x \phantom{.} dx & = (x - 1)e^x \\ \\ \text{Area of shaded region } A & = - \int_{-2}^0 (xe^x) \phantom{0} dx \\ & = - \left[ (x - 1)e^x \right]_{-2}^0 \\ & = - \left\{ \left[ (0 - 1)e^0 \right] - \left[ (-2 - 1)e^{-2} \right] \right\} \\ & = - \left\{ \left[ (-1)(1) \right] - \left[ (-3)e^{-2} \right] \right\} \\ & = -\left( - 1 + 3 e^{-2} \right) \\ & = 0.59399 \\ & \approx 0.594 \text{ units}^2 \\ \\ \\ \text{Area of rectangle} & = 0.5 \times 1 \\ & = 0.5 \text{ units}^2 \\ \\ \text{Area of region } C & = \int_0^{0.5} (xe^x) \phantom{0} dx \\ & = \left[ (x - 1)e^x \right]_0^{0.5} \\ & = \left[ (0.5 - 1)e^{0.5} \right] - \left[ (0 - 1)e^0 \right] \\ & = (-0.5)e^{0.5} - (-1)(1) \\ & = 0.17563 \text{ units}^2 \\ \\ \therefore \text{Area of shaded region } B & = 0.5 - 0.17563 \\ & = 0.32437 \\ & \approx 0.324 \text{ units}^2 \end{align}

(i)

\begin{align} \int_{\pi \over 6}^{\pi \over 2} \sin x \phantom{0} dx & = \left[ {-\cos x \over 1} \right]_{\pi \over 6}^{\pi \over 2} \\ & = \left[ - \cos x \right]_{\pi \over 6}^{\pi \over 2} \\ & = \left[ - \cos {\pi \over 2} \right] - \left[ - \cos {\pi \over 6} \right] \\ & = \left[ - (0) \right] - \left[ - \left( \sqrt{3} \over 2 \right) \right] \\ & = 0 + {\sqrt{3} \over 2} \\ & = {\sqrt{3} \over 2} \end{align}

(ii)

\begin{align} y & = \sin x \\ \\ \text{When } & y = {1 \over 2}, \\ {1 \over 2} & = \sin x \\ \implies x & = {\pi \over 6} \phantom{000000} \left[\text{Special value: } \sin {\pi \over 6} = {1 \over 2} \right] \\ \\ \text{When } & y = 1, \\ 1 & = \sin x \\ \implies x & = {\pi \over 2} \phantom{000000} \left[\text{Special value: } \sin {\pi \over 2} = 1 \right] \end{align}

Region A is a rectangle. Regions B and C combine to form another rectangle.

\begin{align} \text{Area of region } A & = \text{Length } \times \text{Breadth} \\ & = {\pi \over 6} \times {1 \over 2} \\ & = {\pi \over 12} \text{ units}^2 \\ \\ \text{Area of rectangle (} B + C) & = \text{Length } \times \text{Breadth} \\ & = 1 \times \left({\pi \over 2} - {\pi \over 6} \right) \\ & = 1 \times {\pi \over 3} \\ & = {\pi \over 3} \text{ units}^2 \\ \\ \text{Area of region } C & = \int_{\pi \over 6}^{\pi \over 2} \sin x \phantom{0} dx \\ & = {\sqrt{3} \over 2} \text{ units}^2 \phantom{000000} [\text{Use answer from (i)}] \\ \\ \text{Area of region } B & = \left( {\pi \over 3} - {\sqrt{3} \over 2} \right) \text{ units}^2 \\ \\ \text{Area of shaded region} & = {\pi \over 12} + {\pi \over 3} - {\sqrt{3} \over 2} \\ & = {\pi \over 12} + {4 \pi \over 12} - {\sqrt{3} \over 2} \\ & = \left( {5 \pi \over 12} - {\sqrt{3} \over 2} \right) \text{ units}^2 \phantom{00} \text{ (Shown)} \end{align}

(i)

\begin{align} \text{Area of region } A & = \int_0^{\pi \over 4} \sec^2 x \phantom{0} dx \\ & = \left[ \tan x \right]_0^{\pi \over 4} \\ & = \tan {\pi \over 4} - \tan 0 \\ & = 1 - 0 \\ & = 1 \text{ units}^2 \end{align}

(ii) Regions A and B combine to form a trapezium

\begin{align} y & = 1 + {4 \over \pi}x \\ \\ \text{When } & x = {\pi \over 4}, \\ y & = 1 + {4 \over \pi} \left({\pi \over 4} \right) \\ & = 1 + 1 \\ & = 2 \\ \\ \therefore & \phantom{.} P \left( {\pi \over 4}, 2 \right) \\ \\ \text{Area of trapezium } & = {1 \over 2} \times \text{Sum of parallel sides } \times \text{Height} \\ & = {1 \over 2} \times (1 + 2) \times {\pi \over 4} \\ & = {1 \over 2} \times 3 \times {\pi \over 4} \\ & = {3\pi \over 8} \text{ units}^2 \\ \\ \therefore \text{Area of shaded region } B & = {3\pi \over 8} - 1 \\ & = \left( {3\pi \over 8} - 1 \right) \text{ units}^2 \phantom{00} \text{ (Shown)} \end{align}

(i)

\begin{align} y & = e^x \\ \\ \text{When } & x = 2, \\ y & = e^2 \\ \\ \therefore & \phantom{.} A(2, e^2) \\ \\ \\ y & = e^x \\ \\ {dy \over dx} & = e^x \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = e^2 \\ \\ y & = mx + c \phantom{000000} [\text{Form equation of tangent}] \\ y & = e^2 x + c \\ \\ \text{Using } & A(2 , e^2), \\ e^2 & = e^2 (2) + c \\ e^2 & = 2e^2 + c \\ e^2 - 2e^2 & = c \\ - e^2 & = c \\ \\ \text{Eqn of tangent: } & y = e^2 x - e^2 \\ \\ \text{When } & y = 0, \\ 0 & = e^2x - e^2 \\ e^2 & = e^2 x \\ {e^2 \over e^2} & = x \\ 1 & = x \\ \\ \therefore & \phantom{.} B(1, 0) \end{align}

(ii)

\begin{align} \text{Area bounded by curve} & = \int_0^2 e^x \phantom{0} dx \\ & = \left[ {e^x \over 1} \right]_0^2 \\ & = \left[ e^x \right]_0^2 \\ & = e^2 - e^0 \\ & = (e^2 - 1) \text{ units}^2 \\ \\ \text{Area of triangle } ABC & = {1 \over 2} \times BC \times AC \\ & = {1 \over 2} \times (1) \times e^2 \\ & = {1 \over 2} e^2 \\ \\ \text{Area of shaded region} & = e^2 - 1 - {1 \over 2}e^2 \\ & = \left( {1 \over 2}e^2 - 1 \right) \text{ units}^2 \phantom{00} \text{ (Shown)} \end{align}