\begin{align} \text{Area of shaded region } & = \int_0^4 x(4 - x) \phantom{0} dx \\ & = \int_0^4 4x - x^2 \phantom{0} dx \\ & = \left[ {4x^2 \over 2} - {x^3 \over 3} \right]_0^4 \\ & = \left[ 2x^2 - {1 \over 3}x^3 \right]_0^4 \\ & = \left[2(4)^2 - {1 \over 3}(4)^3\right] - \left[2(0)^2 - {1 \over 3}(0)^3\right] \\ & = {32 \over 3} - 0 \\ & = {32 \over 3} \\ & = 10{2 \over 3} \text{ units}^2 \end{align}

\begin{align} \text{Area of shaded region } & = \int_0^{\pi \over 4} \cos 2x \phantom{0} dx \\ & = \left[ {\sin 2x \over 2} \right]_0^{\pi \over 4} \\ & = {\sin [2({\pi \over 4})] \over 2} - {\sin [2(0)] \over 2} \\ & = {\sin {\pi \over 2} \over 2} - {\sin 0 \over 2} \phantom{000000} [\text{Radian mode!}] \\ & = {1 \over 2} - {0 \over 2} \\ & = {1 \over 2} \text{ units}^2 \end{align}

\begin{align} \text{Area of shaded region } & = \int_0^1 e^{2x} \phantom{0} dx \\ & = \left[ {e^{2x} \over 2} \right]_0^1 \\ & = {e^{2(1)} \over 2} - {e^{2(0)} \over 2} \\ & = {e^2 \over 2} - {e^0 \over 2} \\ & = {e^2 \over 2} - {1 \over 2} \\ & = {1 \over 2} (e^2 - 1) \text{ units}^2 \end{align}

\begin{align} \text{Area of shaded region } & = \int_0^3 \left[ (x - 1)^2 + 1 \right] \phantom{0} dx \\ & = \left[ {(x - 1)^3 \over (3)(1)} + x \right]_0^3 \\ & = \left[ {1 \over 3}(x - 1)^3 + x \right]_0^3 \\ & = \left[ {1 \over 3}(3 - 1)^3 + 3 \right] - \left[ {1 \over 3}(0 - 1)^3 + 0 \right] \\ & = \left[ {1 \over 3}(8) + 3 \right] - \left[ {1 \over 3}(-1) \right] \\ & = {8 \over 3} + 3 + {1 \over 3} \\ & = 6 \text{ units}^2 \end{align}

\begin{align} \text{Area of shaded region } & = -\int_0^\pi \sin (x - \pi) \phantom{0} dx \\ & = -\left[ {-\cos (x - \pi) \over 1} \right]_0^\pi \\ & = -\left[ -\cos (x - \pi) \right]_0^\pi \\ & = - \{ \left[ -\cos (\pi - \pi) \right] - \left[ -\cos (0 - \pi) \right] \} \\ & = - [ -\cos 0 + \cos (-\pi) ] \\ & = - [ - (1) + (-1) ] \\ & = - ( - 2) \\ & = 2 \text{ units}^2 \end{align}

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

\begin{align} \text{Area of shaded region } & = \int_1^4 [(y - 3)^2 + 2] \phantom{0} dy \\ & = \left[ {(y - 3)^3 \over (3)(1)} + 2y \right]_1^4 \\ & = \left[ {1 \over 3}(y - 3)^3 + 2y \right]_1^4 \\ & = \left[ {1 \over 3}(4 - 3)^3 + 2(4) \right] - \left[ {1 \over 3}(1 - 3)^3 + 2(1) \right] \\ & = \left[ {1 \over 3}(1) + 8 \right] - \left[ {1 \over 3}(-8) + 2\right] \\ & = \left[ {1 \over 3} + 8 \right] - \left[ -{8 \over 3} + 2 \right] \\ & = 8{1 \over 3} - \left(-{2 \over 3}\right) \\ & = 9 \text{ units}^2 \end{align}

\begin{align} \text{Area of shaded region } & = \int_0^2 y(2 - y) \phantom{0} dy \\ & = \int_0^2 2y - y^2 \phantom{0} dy \\ & = \left[ {2y^2 \over 2} - {y^3 \over 3} \right]_0^2 \\ & = \left[ y^2 - {1 \over 3}y^3 \right]_0^2 \\ & = \left[ (2)^2 - {1 \over 3}(2)^3 \right] - \left[ (0)^2 - {1 \over 3}(0)^3 \right] \\ & = {4 \over 3} - 0 \\ & = 1{1 \over 3} \text{ units}^2 \end{align}

(i)

\begin{align} y & = \sin \left(2x + {\pi \over 3} \right) \\ \\ \text{When } & y = 0, \\ 0 & = \sin \left(2x + {\pi \over 3} \right) \end{align}

\begin{align} 2x + {\pi \over 3} & = 0, \pi, 2\pi \\ \\ 2x & = 0 - {\pi \over 3},\phantom{0} \pi - {\pi \over 3},\phantom{0} 2\pi - {\pi \over 3} \\ & = -{\pi \over 3}, \phantom{0} {2\pi \over 3}, \phantom{0} {5\pi \over 3} \\ \\ x & = -{\pi \over 6}, \phantom{0} {\pi \over 3}, \phantom{0} {5\pi \over 6} \\ \\ \therefore x \text{-coor} & \text{dinate of } A = {\pi \over 3} \end{align}

(ii)

\begin{align} \text{Area of shaded region } & = \int_0^{\pi \over 3} \sin \left(2x + {\pi \over 3} \right) \phantom{0} dx \\ & = \left[ {-\cos (2x +{\pi \over 3}) \over 2} \right]_0^{\pi \over 3} \\ & = \left[ {-\cos (2({\pi \over 3}) + {\pi \over 3}) \over 2} \right] - \left[ {-\cos (2(0) + {\pi \over 3}) \over 2} \right] \\ & = \left[ {-\cos \pi \over 2} \right] - \left[ {-\cos {\pi \over 3} \over 2} \right] \\ & = \left[ {-(-1) \over 2} \right] - \left[ {-({1 \over 2}) \over 2 } \right] \\ & = {1 \over 2} - \left(- {1 \over 4}\right) \\ & = {3 \over 4} \text{ units}^2 \end{align}

(i)

\begin{align} \text{Area of shaded region } & = \int_0^a e^{-x} \phantom{0} dx \\ & = \left[ {e^{-x} \over -1} \right]_0^a \\ & = \left[ -e^{-x} \right]_0^a \\ & = \left[ -e^{-a} \right] - \left[ -e^{0} \right] \\ & = -e^{-a} - [-(1)] \\ & = -e^{-a} + 1 \\ & = (1 - e^{-a}) \text{ units}^2 \end{align}

(ii)

\begin{align} 0.5 & = 1 - e^{-a} \\ -0.5 & = -e^{-a} \\ 0.5 & = e^{-a} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln 0.5 & = \ln e^{-a} \\ \ln 0.5 & = (-a) \ln e \phantom{0000000000} [\text{Power law (logarithms)}] \\ \ln 0.5 & = (-a)(1) \\ \ln 0.5 & = -a \\ -\ln 0.5 & = a \\ \\ a & = -\ln 0.5 \\ & = 0.69314 \\ & \approx 0.693 \end{align}

(a)

\begin{align} \text{Area bounded } & = \int_2^5 x^3 + 1 \phantom{0} dx \\ & = \left[ {x^4 \over 4} + x \right]_2^5 \\ & = \left[ {(5)^4 \over 4} + 5 \right] - \left[ {(2)^4 \over 4} + 2 \right] \\ & = 161{1 \over 4} - 6 \\ & = 155{1 \over 4} \text{ units}^2 \end{align}

(b)

\begin{align} \text{Area bounded } & = \int_1^2 2 - {3 \over x^2} \phantom{0} dx \\ & = \int_1^2 2 - 3x^{-2} \phantom{0} dx \\ & = \left[ 2x - {3x^{-1} \over -1} \right]_1^2 \\ & = \left[ 2x + 3x^{-1} \right]_1^2 \\ & = \left[ 2x + {3 \over x} \right]_1^2 \\ & = \left[ 2(2) + {3 \over 2} \right] - \left[ 2(1) + {3 \over 1} \right] \\ & = 5{1 \over 2} - 5 \\ & = {1 \over 2} \text{ units}^2 \end{align}

The shaded region can be divided into two regions:

Region A is bounded by the curve while Region B is a triangle
\begin{align} \text{Area of }A & = \int_0^1 x^2 + 1 \phantom{0} dx \\ & = \left[ {x^3 \over 3} + x \right]_0^1 \\ & = \left[ {(1)^3 \over 3} + 1 \right] - \left[ {(0)^3 \over 3} + 0 \right] \\ & = {4 \over 3} - 0 \\ & = {4 \over 3} \text{ units}^2 \\ \\ \text{Area of }B & = {1 \over 2} \times \text{Base } \times \text{Height} \\ & = {1 \over 2} \times 3 \times 2 \\ & = 3 \text{ units}^2 \\ \\ \therefore \text{Area of shaded region } & = {4 \over 3} + 3 \\ & = 4{1 \over 3} \text{ units}^2 \end{align}

Note the area bounded by the curve is below the x-axis

\begin{align} y & = (x + 1)(x - 2) \\ & = x^2 - 2x + x - 2 \\ & = x^2 - x - 2 \\ \\ \text{Area of shaded region } & = -\int_{-1}^2 (x^2 - x - 2) \phantom{0} dx \phantom{0} \\ & = -\left[ {x^3 \over 3} - {x^2 \over 2} - 2x \right]_{-1}^2 \\ & = -\left\{ \left[ {(2)^3 \over 3} - {(2)^2 \over 2} - 2(2) \right] - \left[ {(-1)^3 \over 3} - {(-1)^2 \over 2} - 2(-1) \right]\right\} \\ & = -\left[ -3{1 \over 3} - 1{1 \over 6} \right] \\ & = 4{1 \over 2} \text{ units}^2 \end{align}

(i)

Since the coefficient of $x^2$ is positive, this is a minimum quadratic curve (happy face)

\begin{align} y & = x^2 - x - 6 \\ \\ \text{When } & x = 0, \\ y & = (0)^2 - (0) - 6 \\ & = -6 \\ \implies & y \text{-intercept is } (0, - 6) \\ \\ \text{When } & y = 0, \\ 0 & = x^2 - x - 6 \\ 0 & = (x - 3)(x + 2) \\ \\ x - 3 = 0 \phantom{00} & \text{or} \phantom{000} x + 2 = 0 \\ x = 3 \phantom{00} & \phantom{or000+2} x = - 2 \\ \\ \implies & x \text{-intercepts are } (3, 0) \text{ and } (-2, 0) \\ \\ \text{Line of symmetry, } x & = {3 + (-2) \over 2} \\ x & = {1 \over 2} \\ x & = 0.5 \\ \\ \text{When } & x = 0.5, \\ y & = \left(1 \over 2\right)^2 - {1 \over 2} - 6 \\ & = -6{1 \over 4} \\ & = -6.25 \\ \implies & \text{Turning point is } (0.5, -6.25) \end{align}

(ii) Note the area bounded by the curve is below the x-axis

\begin{align} \text{Area bounded } & = - \int_{-2}^3 (x^2 - x - 6) \phantom{0} dx \\ & = - \left[ {x^3 \over 3} - {x^2 \over 2} - 6x \right]_{-2}^3 \\ & = - \left\{ \left[ {(3)^3 \over 3} - {(3)^2 \over 2} - 6(3) \right] - \left[ {(-2)^3 \over 3} - {(-2)^2 \over 2} - 6(-2) \right] \right\} \\ & = -\left[ -13{1 \over 2} - 7{1 \over 3} \right] \\ & = 20{5 \over 6} \text{ units}^2 \end{align}

(i)

\begin{align} y & = x^2 \phantom{0} \text{ --- (1)} \\ \\ x + y & = 6 \\ y & = 6 - x \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 & = 6 - x \\ x^2 + x - 6 & = 0 \\ (x - 2)(x + 3) & = 0 \\ \\ x - 2 = 0 \phantom{00} & \text{or} \phantom{000} x + 3 = 0 \\ x = 2 \phantom{00} & \phantom{or000+3} x = - 3 \text{ (Reject)} \\ \\ x\text{-coordinate of } A & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into (1),} \\ \text{When } & x = 2, \\ y & = (2)^2 \\ & = 4 \\ \\ \therefore & \phantom{.} A(2, 4) \end{align}

(ii)

A trapezium is enclosed by the $y$-axis, the line $x + y = 6$, the vertical line drawn from A to the $x$-axis and the $x$-axis.

Thus, the area of the shaded region can be calculated by subtracting the area of the unshaded region ('B') from the area of the trapezium.

\begin{align} \text{Area of trapezium } & = {1 \over 2} \times \text{Sum of parallel sides } \times \text{Height } \\ & = {1 \over 2} \times (6 + 4) \times 2 \\ & = 10 \text{ units}^2 \\ \\ \text{Area of unshaded region } B & = \int_0^2 x^2 \phantom{0} dx \\ & = \left[ {x^3 \over 3} \right]_0^2 \\ & = {(2)^3 \over 3} - {(0)^3 \over 3} \\ & = {8 \over 3} - 0 \\ & = {8 \over 3} \text{ units}^2 \\ \\ \therefore \text{Area of shaded region } & = \text{Area of trapezium } - \text{Area of } B \\ & = 10 - {8 \over 3} \\ & = 7{1 \over 3} \text{ units}^2 \end{align}

(i)

\begin{align} x & = 4y - y^2 \phantom{0} \text{ ---- (1)} \\ \\ x & = 4 - y \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4y - y^2 & = 4 - y \\ -y^2 + 5y - 4 & = 0 \\ y^2 - 5y + 4 & = 0 \\ (y - 4)(y - 1) & = 0 \\ \\ y - 4 = 0 \phantom{00} & \text{or} \phantom{000} y - 1 = 0 \\ y = 4 \phantom{00} & \phantom{or000-1} y = 1 \\ \\ \text{Substitute } & y = 4 \text{ into (2),} \\ x = 4 - 4 \phantom{00} & \phantom{00000} x = 4 - 1 \\ x = 0 \phantom{-4000} & \phantom{00000} x = 3 \\ \\ \therefore A(0, 4) & \text{ and } B(3, 1) \end{align}

(ii)

The shaded region can be divided into two regions, C and D. Since AB is part of a straight line, region C is a right-angled triangle.

Region D is bounded the $y$-axis, horizontal line from B to the $y$-axis and the curve $x = 4y - y^2$.

\begin{align} \text{Area of }C & = {1 \over 2} \times \text{Base } \times \text{Height } \\ & = {1 \over 2} \times 3 \times 3 \\ & = 4{1 \over 2} \text{ units}^2 \\ \\ \text{Area of }D & = \int_0^1 (4y - y^2) \phantom{0} dy \phantom{000000} [\text{w.r.t } y \text{-axis}] \\ & = \left[ {4y^2 \over 2} - {y^3 \over 3} \right]_0^1 \\ & = \left[ 2y^2 - {y^3 \over 3} \right]_0^1 \\ & = \left[ 2(1)^2 - {(1)^3 \over 3} \right] - \left[ 2(0)^2 - {(0)^3 \over 3} \right] \\ & = 1{2 \over 3} - 0 \\ & = 1{2 \over 3} \text{ units}^2 \\ \\ \therefore \text{Area of shaded region} & = 4{1 \over 2} + 1{2 \over 3} \\ & = 6{1 \over 6} \text{ units}^2 \end{align}

(i)

\begin{align} {d \over dx} (\sqrt{3x + 7}) & = {d \over dx} (3x + 7)^{1 \over 2} \\ & = {1 \over 2}(3x + 7)^{-{1 \over 2}} .(3) \phantom{000000} [\text{Chain rule}] \\ & = {3 \over 2}(3x + 7)^{-{1 \over 2}} \\ & = {3 \over 2}\left(1 \over \sqrt{3x + 7}\right) \\ & = {3 \over 2\sqrt{3x + 7}} \end{align}

(ii)

\begin{align} \text{From (i), } {d \over dx} (\sqrt{3x + 7}) & = {3 \over 2\sqrt{3x + 7}} \\ \\ \implies \int {3 \over 2\sqrt{3x + 7}} \phantom{.} dx & = \sqrt{3x + 7} \\ \\ \text{Area of shaded region} & = \int_{-1}^3 {1 \over \sqrt{3x + 7}} \phantom{.} dx \\ & = {2 \over 3} \int_{-1}^3 {3 \over 2\sqrt{3x + 7}} \phantom{.} dx \\ & = {2 \over 3} \left[\sqrt{3x + 7}\right]_{-1}^3 \\ & = {2 \over 3} \left[ \sqrt{3(3) + 7} - \sqrt{3(-1) + 7} \right] \\ & = {2 \over 3} (4 - 2) \\ & = {4 \over 3} \text{ units}^2 \end{align}

(i)

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

(ii)

\begin{align} y & = 1 + x^2 \\ \\ \text{When } & x = a, \\ y & = 1 + (a)^2 \\ & = 1 + a^2 \\ \text{Coordinates } & \text{of point of intersection is } (a , 1 + a^2) \\ \\ \text{Area of rectangle} & = \text{Length } \times \text{Breadth} \\ & = (1 + a^2) \times a \\ & = (1 + a^2)(a) \\ & = (a + a^3) \\ \\ \therefore \text{Area of shaded region} B & = \text{Area of rectangle } - \text{Area of shaded region }A \\ & = (a + a^3) - \left(a + {a^3 \over 3} \right) \\ & = a + a^3 - a - {a^3 \over 3} \\ & = {2a^3 \over 3} \\ \\ \text{Area of } B & = \text{Area of } A \\ {2a^3 \over 3} & = a + {a^3 \over 3} \\ {2a^3 \over 3} - {a^3 \over 3} - a & = 0 \\ {a^3 \over 3} - a & = 0 \\ a^3 - 3a & = 0 \\ a(a^2 - 3) & = 0 \end{align} \begin{align} a & = 0 \text{ (Reject, } a > 0) && \text{ or } & a^2 - 3 & = 0 \\ & &&& a^2 & = 3 \\ & &&& a & = \pm \sqrt{3} \\ & &&& a & = \sqrt{3} \phantom{.} \text{ or } \phantom{.} - \sqrt{3} \text{ (Reject, } a > 0) \end{align}
$$ \therefore a = \sqrt{3} \approx 1.73 \text{ (2 d.p.)} $$

\begin{align} y & = {10 \over x^2} \\ & = 10x^{-2} \\ \\ \int 10x^{-2} \phantom{0} dx & = {10x^{-1} \over -1} \\ & = -10x^{-1} \\ & = -{10 \over x} \\ \\ \text{Area of region }P & = \int_2^k 10x^{-2} \phantom{0} dx \\ & = \left[-{10 \over x} \right]_2^k \\ & = \left[-{10 \over k} \right] - \left[-{10 \over 2}\right] \\ & = -{10 \over k} - (-5) \\ & = -{10 \over k} + 5 \\ \\ \text{Area of region }Q & = \int_k^5 10x^{-2} \phantom{0} dx \\ & = \left[-{10 \over x} \right]_k^5 \\ & = \left[-{10 \over 5} \right] - \left[-{10 \over k}\right] \\ & = -2 + {10 \over k} \\ \\ \text{Area of } P & = \text{Area of } Q \\ -{10 \over k} + 5 & = -2 + {10 \over k} \\ -{10 \over k} - {10 \over k} & = -2 - 5 \\ -{20 \over k} & = - 7 \\ {20 \over k} & = 7 \\ 20 & = 7k \\ {20 \over 7} & = k \end{align}

(i)

\begin{align} {dy \over dx} & = {d \over dx}(x^2 - ax - b) \\ & = 2x - a - 0 \\ & = 2x - a \\ \\ \text{When } x = 1 & \text{ and } {dy \over dx} = 0, \phantom{000000} [\text{Minimum point } (1, -4)] \\ 0 & = 2(1) - a \\ 0 & = 2 - a \\ a & = 2 \\ \\ \text{Eqn of curve: } & y = x^2 - 2x - b \\ \\ \text{Using } & (1, -4), \\ -4 & = (1)^2 - 2(1) - b \\ -4 & = 1 - 2 - b \\ -3 & = - b \\ 3 & = b \\ \\ \therefore a & = 2, b = 3 \end{align}

(ii) Note $P$ and $Q$ are the x-intercepts while $R$ is the y-intercept

\begin{align} \text{Eqn of curve: } & y = x^2 - 2x - 3 \\ \\ \text{Let } & y = 0, \\ 0 & = x^2 - 2x - 3 \\ 0 & = (x - 3)(x + 1) \\ \\ x - 3 = 0 \phantom{00} & \text{or} \phantom{000} x + 1 = 0 \\ x = 3 \phantom{00} & \phantom{or000+1} x = -1 \\ \\ \\ \therefore & \phantom{.} P(-1, 0) \text{ and } Q(3, 0) \\ \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 - 2(0) - 3 \\ & = 0 - 0 - 3 \\ & = -3 \\ \\ \therefore & \phantom{.} R(0, -3) \end{align}

(iii)

\begin{align} \text{Area of bounded by curve} & = - \int_{-1}^3 (x^2 - 2x - 3) dx \\ & = - \left[ {x^3 \over 3} - {2x^2 \over 2} - 3x \right]_{-1}^3 \\ & = - \left[ {x^3 \over 3} - x^2 - 3x \right]_{-1}^3 \\ & = -\left\{ \left[ {(3)^3 \over 3} - (3)^2 - 3(3) \right] - \left[ {(-1)^3 \over 3} - (-1)^2 - 3(-1) \right] \right\} \\ & = -\left[ -9 - 1{2 \over 3} \right] \\ & = 10{2 \over 3} \text{ units}^2 \\ \\ \text{Area of } \triangle PQR & = {1 \over 2} \times \text{Base} \times \text{Height} \\ & = {1 \over 2} \times PQ \times OR \\ & = {1 \over 2} \times 4 \times 3 \\ & = 6 \text{ units}^2 \\ \\ \therefore \text{Area of shaded region } & = 10{2 \over 3} - 6 \\ & = 4{2 \over 3} \text{ units}^2 \end{align}

(i)

\begin{align} y & = kx - x^2 \\ \\ \text{Using } & P(2, 0), \\ 0 & = k(2) - (2)^2 \\ 0 & = 2k - 4 \\ 4 & = 2k \\ {4 \over 2} & = k \\ 2 & = k \end{align}

(ii)

\begin{align} y & = 2x - x^2 \\ \\ \text{Using } & Q(3, h), \\ h & = 2(3) - (3)^2 \\ & = 6 - 9 \\ & = -3 \\ \\ \therefore & \phantom{.} Q(3, -3) \end{align}

The shaded region can be divided into region A and region B.

Region C is the region bounded by the curve $y = 2x - x^2$ from $x = 2$ to $x = 3$. The 'sum' of regions B and C form a right-angled triangle.

\begin{align} \text{Area of region } A & = \int_0^2 (2x - x^2) \phantom{0} dx \\ & = \left[ {2x^2 \over 2} - {x^3 \over 3} \right]_0^2 \\ & = \left[ x^2 - {x^3 \over 3} \right]_0^2 \\ & = \left[ (2)^2 - {(2)^3 \over 3} \right] - \left[ (0)^2 - {(0)^3 \over 3}\right] \\ & = 1{1 \over 3} - 0 \\ & = 1{1 \over 3} \text{ units}^2 \\ \\ \\ \text{Area of triangle } & = {1 \over 2} \times 3 \times 3 \\ & = 4{1 \over 2} \text{ units}^2 \\ \\ \text{Area of region } C & = - \int_2^3 (2x - x^2) \phantom{0} dx \\ & = - \left[ x^2 - {x^3 \over 3} \right]_2^3 \\ & = - \left\{ \left[ (3)^2 - {(3)^3 \over 3} \right] - \left[ (2)^2 - {(2)^3 \over 3} \right] \right\} \\ & = - \left[ 0 - 1{1 \over 3} \right] \\ & = 1{1 \over 3} \text{ units}^2 \\ \\ \therefore \text{Area of region } B & = 4{1 \over 2} - 1{1 \over 3} \\ & = 3{1 \over 6} \text{ units}^2 \\ \\ \therefore \text{Area enclosed} & = \text{Area of region } A + \text{Area of region } B \\ & = 1{1 \over 3} + 3{1 \over 6} \\ & = 4{1 \over 2} \text{ units}^2 \end{align}

(a)

\begin{align} \int_0^{2\pi} \sin x \phantom{.} dx & = \int_0^{\pi} \sin x \phantom{.} dx + \int_{\pi}^{2\pi} \sin x \phantom{.} dx \\ & = \int_0^{\pi} \sin x \phantom{.} dx + \left[ - \int_0^{\pi} \sin x \phantom{.} dx \right] \\ & = 0 \end{align}

(b)

\begin{align} a & = \pi, 2\pi, 3 \pi, ... \\ & = k \pi, \text{ where } k \text{ is a positive integer} \end{align}