From P, draw a vertical line down to the x-axis. From Q, draw another vertical line down to the x-axis to form a rectangle.

\begin{align} y & = 8 \phantom{0} \text{ --- (1)} \\ \\ y & = x^2 - 8x + 20 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 8 & = x^2 - 8x + 20 \\ 0 & = x^2 - 8x + 20 - 8 \\ 0 & = x^2 - 8x + 12 \\ 0 & = (x - 2)(x - 6) \\ \\ x - 2 = 0 \phantom{00} & \text{or} \phantom{000} x - 6 = 0 \\ x = 2 \phantom{00} & \phantom{or000-6} x = 6 \\ \\ \therefore & \phantom{.} P(2, 8) \text{ and } Q(6, 8) \\ \\ \text{Area of rectangle} & = 4 \times 8 \\ & = 32 \text{ units}^2 \\ \\ \text{Area under curve} & = \int_2^6 x^2 - 8x + 20 \phantom{0} dx \\ & = \left[ {x^3 \over 3} - {8x^2 \over 2} + 20x \right]_2^6 \\ & = \left[ {1 \over 3}x^3 - 4x^2 + 20x \right]_2^6 \\ & = \left[{1 \over 3}(6)^3 - 4(6)^2 + 20(6) \right] - \left[ {1 \over 3}(2)^3 - 4(2)^2 + 20(2) \right] \\ & = 21{1 \over 3} \text{ units}^2 \\ \\ \text{Area of shaded region} & = 32 - 21{1 \over 3} \\ & = 10{2 \over 3} \text{ units}^2 \end{align}

From P, draw a vertical line down to the x-axis. From Q, draw another vertical line down to the x-axis to form a rectangle.

\begin{align} y & = 8x - x^2 \phantom{0} \text{ --- (1)} \\ \\ y & = 7 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 8x - x^2 & = 7 \\ -x^2 + 8x - 7 & = 0 \\ x^2 - 8x + 7 & = 0 \\ (x - 7)(x - 1) & = 0 \\ \\ x - 7 = 0 \phantom{00} & \text{or} \phantom{000} x - 1 = 0 \\ x = 7 \phantom{00} & \phantom{or000-1} x = 1 \\ \\ \therefore & \phantom{.} P(1, 7) \text{ and } Q(7, 7) \\ \\ \text{Area under curve} & = \int_1^7 8x - x^2 \phantom{.} dx \\ & = \left[{8x^2 \over 2} - {x^3 \over 3}\right]_1^7 \\ & = \left[4x^2 - {1 \over 3}x^3 \right]_1^7 \\ & = \left[4(7)^2 - {1 \over 3}(7)^3\right] - \left[4(1)^2 - {1 \over 3}(1)^3\right] \\ & = 78 \text{ units}^2 \\ \\ \text{Area of rectangle} & = 6 \times 7 \\ & = 42 \text{ units}^2 \\ \\ \text{Area of shaded region} & = 78 - 42 \\ & = 36 \text{ units}^2 \end{align}

From Q, draw a vertical line down to the x-axis to form a rectangle.

\begin{align} \text{From diagram, } & \phantom{.} P \left(0, {1 \over 2}\right) \\ \\ y & = \cos x \phantom{0} \text{ --- (1)} \\ y & = {1 \over 2} \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \cos x & = {1 \over 2} \phantom{000000} [\text{1st & 4th quadrants since } \cos x > 0 ] \end{align}

\begin{align} \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = {\pi \over 3} \\ \\ x & = {\pi \over 3} \phantom{0} , 2\pi - {\pi \over 3} \\ & = {\pi \over 3} \phantom{0} , {5\pi \over 3} \text{ (N.A.)} \\ \\ \therefore & \phantom{.} Q\left({\pi \over 3}, {1 \over 2}\right) \\ \\ \text{Area under curve} & = \int_0^{\pi \over 3} \cos x \phantom{.} dx \\ & = \left[ \sin x \right]_0^{\pi \over 3} \\ & = \sin {\pi \over 3} - \sin 0 \\ & = {\sqrt{3} \over 2} - 0 \\ & = {\sqrt{3} \over 2} \text{ units}^2 \\ \\ \text{Area of rectangle} & = {\pi \over 3} \times {1 \over 2} \\ & = {\pi \over 6} \text{ units}^2 \\ \\ \text{Area of shaded region} & = \left( {\sqrt{3} \over 2} - {\pi \over 6} \right) \text{ units}^2 \end{align}

From P, draw a vertical line down to the x-axis. From Q, draw another vertical line down to the x-axis to form a trapezium.

\begin{align} y & = x^2 -12x + 42 \phantom{0} \text{ --- (1)} \\ \\ y & = x + 2 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 - 12x + 42 & = x + 2 \\ x^2 - 12x - x + 42 - 2 & = 0 \\ x^2 - 13x + 40 & = 0 \\ (x - 8)(x - 5) & = 0 \end{align} \begin{align} x - 8 & = 0 && \text{ or } & x - 5 & = 0 \\ x & = 8 &&& x & = 5 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = 8 + 2 &&& y & = 5 + 2 \\ & = 10 &&& & = 7 \\ \\ \therefore &Q(8, 10) &&& \therefore & P(5, 7) \end{align} \begin{align} \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (7 + 10) \times 3 \\ & = 25{1 \over 2} \text{ units}^2 \\ \\ \text{Area under curve} & = \int_5^8 x^2 - 12x + 42 \phantom{.} dx \\ & = \left[{x^3 \over 3} - {12x^2 \over 2} + 42x \right]_5^8 \\ & = \left[ {1 \over 3}x^3 - 6x^2 + 42x \right]_5^8 \\ & = \left[ {1 \over 3}(8)^3 - 6(8)^2 + 42(8) \right] - \left[ {1 \over 3}(5)^3 - 6(5)^2 + 42(5) \right] \\ & = 21 \text{ units}^2 \\ \\ \text{Area of shaded region} & = 25{1 \over 2} - 21 \\ & = 4{1 \over 2} \text{ units}^2 \end{align}

\begin{align} y & = x + 3 \phantom{0} \text{ --- (1)} \\ \\ y & = -x^2 + 8x - 7 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x + 3 & = -x^2 + 8x - 7 \\ x^2 + x - 8x + 3 + 7 & = 0 \\ x^2 - 7x + 10 & = 0 \\ (x - 5)(x - 2) & = 0 \\ \\ x - 5 = 0 \phantom{00} & \text{or} \phantom{000} x - 2 = 0 \\ x = 5 \phantom{00} & \phantom{or000-3} x = 2 \\ \\ \text{Area under curve} & = \int_2^5 - x^2 + 8x - 7 \phantom{.} dx \\ & = \left[ -{x^3 \over 3} + {8x^2 \over 2} - 7x \right]_2^5 \\ & = \left[ -{1 \over 3}x^3 + 4x^2 - 7x \right]_2^5 \\ & = \left[ -{1 \over 3}(5)^3 + 4(5)^2 - 7(5) \right] - \left[ -{1 \over 3}(2)^3 + 4(2)^2 - 7(2) \right] \\ & = 24 \text{ units}^2 \\ \\ \text{Area under line} & = \int_2^5 x + 3 \phantom{.} dx \phantom{000000} [\text{This is the trapezium}] \\ & = \left[ {x^2 \over 2} + 3x \right]_2^5 \\ & = \left[ {(5)^2 \over 2} + 3(5) \right] - \left[ {(2)^2 \over 2} + 3(2) \right] \\ & = 19{1 \over 2} \text{ units}^2 \\ \\ \text{Area of shaded region} & = 24 - 19{1 \over 2} \\ & = 4{1 \over 2} \text{ units}^2 \end{align}

Shaded region $A$ is bounded by the curve $y = e^x$ and the line $y = 1$ from $x = 0$ to $x = 1$.

Shaded region $B$ is bounded by the line $y = 1$ and the curve $y = 1 - x^2$ from $x = 0$ to $x = 1$.

\begin{align} \text{Area of square} & = 1 \times 1 \\ & = 1 \text{ units}^2 \\ \\ \text{Area under } y = e^x & = \int_0^1 e^x \phantom{.} dx \\ & = [e^x]_0^1 \\ & = e^1 - e^0 \\ & = (e - 1) \text{ units}^2 \\ \\ \text{Area of region } A & = (e - 1) - 1 \\ & = (e - 2) \text{ units}^2 \\ \\ \\ \text{Area under } y = 1 - x^2 & = \int_0^1 1 - x^2 \phantom{.} dx \\ & = \left[x - {x^3 \over 3}\right]_0^1 \\ & = \left[1 - {(1)^3 \over 3}\right] - \left[0 - {(0)^3 \over 3}\right] \\ & = {2 \over 3} \text{ units}^2 \\ \\ \text{Area of region } B & = 1 - {2 \over 3} \\ & = {1 \over 3} \text{ units}^2 \end{align}

(i) Points $P$ and $Q$ are the $x$-intercepts. Point $R$ is the $y$-intercept and point $S$ has the same $y$-coordinate as $R$.

\begin{align} y & = 7 + 6x - x^2 \\ \\ \text{Let } & y = 0, \\ 0 & = 7 + 6x - x^2 \\ 0 & = x^2 - 6x - 7 \\ 0 & = (x - 7)(x + 1) \\ \\ x - 7 = 0 \phantom{00} & \text{or} \phantom{000} x + 1 = 0 \\ x = 7 \phantom{00} & \phantom{or000+1} x = - 1 \\ \\ \therefore & \phantom{.} Q(-1, 0) \text{ and } P(7,0) \\ \\ \\ \text{Let } & x = 0, \\ y & = 7 + 6(0) - (0)^2 \\ & = 7 \\ \\ \therefore & \phantom{.} R(0, 7) \\ \\ \\ \text{Substitute } & y = 7 \text{ into eqn of curve,} \\ 7 & = 7 + 6x - x^2 \\ 0 & = -x^2 + 6x + 7 - 7 \\ 0 & = -x^2 + 6x \\ 0 & = -x(x - 6) \\ \\ -x = 0 \phantom{00} & \text{or} \phantom{000} x - 6 = 0 \\ x = 0 \phantom{00} & \phantom{or000-6} x = 6 \\ \\ \therefore & \phantom{.} S(6, 7) \end{align}

(ii) I will be finding the area of the shaded regions from left to right

\begin{align} \text{Area of first shaded region} & = -\int_{-2}^{-1} (7 + 6x - x^2) \phantom{0} dx \\ & = - \left[ 7x + {6x^2 \over 2} - {x^3 \over 3} \right]_{-2}^{-1} \\ & = - \left[ 7x + 3x^2 - {x^3 \over 3} \right]_{-2}^{-1} \\ & = - \left\{ \left[ 7(-1) + 3(-1)^2 - {(-1)^3 \over 3} \right] - \left[ 7(-2) + 3(-2)^2 - {(-2)^3 \over 3} \right] \right\} \\ & = - \left[ -3{2 \over 3} - {2 \over 3} \right] \\ & = 4{1 \over 3} \text{ units}^2 \\ \\ \\ \text{Area of second shaded region} & = \int_{-1}^0 (7 + 6x - x^2) \phantom{0} dx \\ & = \left[ 7x + {6x^2 \over 2} - {x^3 \over 3} \right]_{-1}^0 \\ & = \left[ 7x + 3x^2 - {x^3 \over 3} \right]_{-1}^0 \\ & = \left[ 7(0) + 3(0)^2 - {(0)^3 \over 3} \right] - \left[ 7(-1) + 3(-1)^2 - {(-1)^3 \over 3} \right] \\ & = 0 - \left(-3{2 \over 3} \right) \\ & = 3{2 \over 3} \text{ units}^2 \\ \\ \\ \text{Area under curve} & = \int_0^6 7 + 6x - x^2 \phantom{.} dx \\ & = \left[ 7x + {6x^2 \over 2} - {x^3 \over 3} \right]_0^6 \\ & = \left[ 7x + 3x^2 - {x^3 \over 3} \right]_0^6 \\ & = \left[ 7(6) + 3(6)^2 - {(6)^3 \over 3} \right] - \left[ 7(0) + 3(0)^2 - {(0)^3 \over 3} \right] \\ & = 78 \text{ units}^2 \\ \\ \text{Area of rectangle } QPSR & = 6 \times 7 \\ & = 42 \text{ units}^2 \\ \\ \text{Area of third shaded region} & = 78 - 42 \\ & = 36 \text{ units}^2 \end{align}

(i)

\begin{align} y & = x^2 - 7x + 15 \phantom{0} \text{ --- (1)} \\ \\ y & = 7 - x \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 - 7x + 15 & = 7 - x \\ x^2 - 7x + x + 15 - 7 & = 0 \\ x^2 - 6x + 8 & = 0 \\ (x - 4)(x - 2) & = 0 \\ \\ x - 4 = 0 \phantom{00} & \text{or} \phantom{000} x - 2 = 0 \\ x = 4 \phantom{00} & \phantom{or000-2} x = 2 \\ \\ x \text{-coordinates of } & \text{point of intersections are } 2 \text{ and } 4 \\ \\ \\ \text{Area under curve} & = \int_1^2 x^2 - 7x + 15 \phantom{.} dx \\ & = \left[{x^3 \over 3} - {7x^2 \over 2} + 15x \right]_1^2 \\ & = \left[ {(2)^3 \over 3} - {7(2)^2 \over 2} + 15(2) \right] - \left[ {(1)^3 \over 3} - {7(1)^2 \over 2} + 15(1) \right] \\ & = 18{2 \over 3} - 11{5 \over 6} \\ & = 6{5 \over 6} \text{ units}^2 \\ \\ \text{Area under line} & = \int_1^2 7 - x \phantom{.} dx \\ & = \left[ 7x - {x^2 \over 2}\right]_1^2 \\ & = \left[7(2) - {(2)^2 \over 2}\right] - \left[7(1) - {(1)^2 \over 2}\right] \\ & = 12 - 6{1 \over 2} \\ & = 5{1 \over 2} \text{ units}^2 \\ \\ \text{Area of region } A & = 6{5 \over 6} - 5{1 \over 2} \\ & = 1{1 \over 3} \text{ units}^2 \end{align}

(ii)

\begin{align} \text{Area under line} & = \int_2^4 7 - x \phantom{.} dx \\ & = \left[ 7x - {x^2 \over 2}\right]_2^4 \\ & = \left[7(4) - {(4)^2 \over 2}\right] - \left[7(2) - {(2)^2 \over 2}\right] \\ & = 20 - 12 \\ & = 8 \text{ units}^2 \\ \\ \text{Area under curve} & = \int_2^4 x^2 - 7x + 15 \phantom{.} dx \\ & = \left[{x^3 \over 3} - {7x^2 \over 2} + 15x \right]_2^4 \\ & = \left[ {(4)^3 \over 3} - {7(4)^2 \over 2} + 15(4) \right] - \left[ {(2)^3 \over 3} - {7(2)^2 \over 2} + 15(2) \right] \\ & = 25{1 \over 3} - 18{2 \over 3} \\ & = 6{2 \over 3} \text{ units}^2 \\ \\ \text{Area of region } B & = 8 - 6{2 \over 3} \\ & = 1{1 \over 3} \text{ units}^2 \end{align}

(i)

\begin{align} x \text{-coordinate of } P & = 3 \\ \\ \text{Substitute } & x = 3 \text{ into eqn of curve,} \\ y & = (3)^2 + 4(3) - 5 \\ & = 16 \\ \\ \therefore & \phantom{.} P(3, 16) \end{align}

(ii)

\begin{align} y & = x^2 + 4x - 5 \\ \\ \text{Let } & y = 0, \\ 0 & = x^2 + 4x - 5 \\ 0 & = (x - 1)(x + 5) \\ \\ x - 1 = 0 \phantom{00} & \text{or} \phantom{000} x + 5 = 0 \\ x = 1 \phantom{00} & \phantom{or000+5} x = - 5 \\ \\ \therefore & \phantom{.} R(1, 0) \\ \\ \text{Area under curve} & = \int_1^3 x^2 + 4x - 5 \phantom{.} dx \\ & = \left[ {x^3 \over 3} + {4x^2 \over 2} - 5x \right]_1^3 \\ & = \left[ {1 \over 3}x^3 + 2x^2 - 5x \right]_1^3 \\ & = \left[ {1 \over 3}(3)^3 + 2(3)^2 - 5(3) \right] - \left[ {1 \over 3}(1)^3 + 2(1)^2 - 5(1) \right] \\ & = 12 - \left(-2{2 \over 3}\right) \\ & = 14{2 \over 3} \text{ units}^2 \\ \\ \\ y & = mx + c \phantom{000000} [\text{Form eqn of tangent at } P(3, 16)] \\ y & = 10x + c \\ \\ \text{Using } & P(3, 16), \\ 16 & = 10(3) + c \\ 16 & = 30 + c \\ -14 & = c \\ \\ \text{Eqn of tangent: } & y = 10x - 14 \\ \\ \text{Let } & y = 0, \\ 0 & = 10x - 14 \\ 14 & = 10x \\ {14 \over 10} & = x \\ 1{2 \over 5} & = x \\ \\ \therefore & \phantom{.} Q \left(1{2 \over 5}, 0 \right) \\ \\ \text{Area of triangle} & = {1 \over 2} \times \left(3 - 1{2 \over 5}\right) \times 16 \\ & = 12{4 \over 5} \text{ units}^2 \\ \\ \text{Area of shaded region} & = 14{2 \over 3} - 12{4 \over 5} \\ & = 1{13 \over 15} \text{ units}^2 \end{align}

(i) Point $A$ is the $y$-intercept of both the curve and the line.

\begin{align} y & = 7x + 6x - x^2 \\ \\ \text{Let } & x = 0, \\ y & = 7 + 6(0) - (0)^2 \\ & = 7 \\ \\ \therefore & \phantom{.} A(0, 7) \\ \\ y & = mx + c \\ \\ \text{Using } & A(0, 7), \\ 7 & = m(0) + c \\ 7 & = 0 + c \\ 7 & = c \end{align}

(ii)

\begin{align} y & = mx + c \\ y & = mx + 7 \phantom{0} \text{ --- (1)} \\ \\ y & = 7 + 6x - x^2 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ mx + 7 & = 7 + 6x - x^2 \\ x^2 + mx - 6x + 7 - 7 & = 0 \\ x^2 + mx - 6x & = 0 \\ x(x + m - 6) & = 0 \\ \\ x = 0 \phantom{00} & \text{or} \phantom{000} x + m - 6 = 0 \\ & \phantom{or000+m-6} x = 6 - m \\ \\ x\text{-coordinate of } B & = 6 - m \end{align}

(iii)

\begin{align} \text{Area under curve} & = \int_0^{6 - m} 7 + 6x - x^2 \phantom{.} dx \\ & = \left[7x + {6x^2 \over 2} - {x^3 \over 3} \right]_0^{6 - m} \\ & = \left[7x + 3x^2 - {1 \over 3}x^3 \right]_0^{6 - m} \\ & = \left[7(6 - m) + 3(6 - m)^2 - {1 \over 3}(6 - m)^3\right] - \left[ 7(0) + 3(0)^2 - {1 \over 3}(0)^3 \right] \\ & = 7(6 - m) + 3(6 - m)^2 - {1 \over 3}(6 - m)^3 - 0 \\ & = 7(6 - m) + 3(6 - m)^2 - {1 \over 3}(6 - m)^3 \\ \\ \text{Area under line} & = \int_0^{6 - m} mx + 7 \phantom{.} dx \\ & = \left[ {mx^2 \over 2} + 7x \right]_0^{6 - m} \\ & = \left[ {m \over 2} x^2 + 7x \right]_0^{6 - m} \\ & = \left[ {m \over 2} (6 - m)^2 + 7(6 - m) \right] - \left[ {m \over 2}(0)^2 + 7(0) \right] \\ & = {m \over 2} (6 - m)^2 + 7(6 - m) - 0 \\ & = {m \over 2} (6 - m)^2 + 7(6 - m) \\ \\ \text{Area of shdaed region} & = 7(6 - m) + 3(6 - m)^2 - {1 \over 3}(6 - m)^3 - \left[ {m \over 2} (6 - m)^2 + 7(6 - m) \right] \\ & = 7(6 - m) + 3(6 - m)^2 - {1 \over 3}(6 - m)^3 - {m \over 2}(6 - m)^2 - 7(6 - m) \\ & = 3(6 - m)^2 - {m \over 2}(6 - m)^2 - {1 \over 3}(6 - m)^3 \\ & = (6 - m)^2 \left[ 3 - {m \over 2} - {1 \over 3}(6 - m) \right] \\ & = (6 - m)^2 \left( 3 - {1 \over 2}m - 2 + {1 \over 3}m \right) \\ & = (6 - m)^2 \left( 1 - {1 \over 6}m \right) \\ \\ 20{5 \over 6} & = (6 - m)^2 \left( 1 - {1 \over 6}m \right) \\ 6 \left(20{5 \over 6} \right) & = 6 \left[(6 - m)^2 \left( 1 - {1 \over 6}m \right)\right] \\ 125 & = (6 - m)^2 \left[6 \left( 1 - {1 \over 6}m \right)\right] \\ 125 & = (6 - m)^2 (6 - m) \\ 125 & = (6 - m)^3 \\ \sqrt[3]{125} & = 6 - m \\ 5 & = 6 - m \\ m & = 6 - 5 \\ m & = 1 \end{align}

(i) The line passes through the origin $(0, 0)$ and point $Q (1, 1)$.

\begin{align} \text{Gradient } & = {y_2 - y_1 \over x_2 - x_1} \\ & = {1 - 0 \over 1 - 0} \\ & = 1 \\ \\ y & = mx + c \\ y & = x + c \\ \\ \text{Using } & Q(1, 1), \\ 1 & = 1 + c \\ 0 & = c \\ \\ \text{Eqn of line: } & y = x \end{align}

(ii)

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

(iii) Shaded regions $A$ and $B$ form a trapezium.

\begin{align} \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} (1 + k ) (k - 1) \\ & = {1 \over 2}(k + 1)(k - 1) \\ & = {(k + 1)(k - 1) \over 2} \\ \\ \text{Area of shaded region } B & = \text{Area of trapezium} - \text{Area of shaded region } A \\ & = {(k + 1)(k - 1) \over 2} - {k - 1 \over k} \\ & = {k(k + 1)(k - 1) \over 2k} - {2(k - 1) \over 2k} \\ & = {k(k + 1)(k - 1) - 2(k - 1) \over 2k} \\ & = {( k - 1) [ k(k + 1) - 2] \over 2k} \\ & = {(k - 1)(k^2 + k - 2) \over 2k} \phantom{00} \text{ (Shown)} \end{align}

\begin{align} \text{Finite area under the graph from } x \ge 1 \end{align}