(i)

\begin{align} {d \over dx}(x^2 + 5x) & = 2x + 5(1) \\ & = 2x + 5 \end{align}

(ii)

\begin{align} \text{Since } {d \over dx}(x^2 + 5x) & = 2x + 5, \\ \int (2x + 5) \phantom{.} dx & = x^2 + 5x + c \end{align}

(i)

\begin{align} u & = x &&& v & = 1 + 2x \\ {du \over dx} & = 1 &&& {dv \over dx} & = 2(1) \\ & &&& & = 2 \end{align} \begin{align} {d \over dx} \left(x \over 1 + 2x\right) & = { (1 + 2x)(1) - (x)(2) \over (1 + 2x)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = {1 + 2x - 2x \over (1 + 2x)^2} \\ & = {1 \over (1 + 2x)^2} \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{Since } {d \over dx} \left(x \over 1 + 2x\right) & = {1 \over (1 + 2x)^2}, \\ \int {1 \over (1 + 2x)^2 } \phantom{.} dx & = {x \over 1 + 2x} \\ \\ \therefore \int {2 \over (1 + 2x)^2 } \phantom{.} dx & = 2 \int {1 \over (1 + 2x)^2 } \phantom{.} dx \\ & = 2 \left(x \over 1 + 2x\right) \\ & = {2 \over 1} \left(x \over 1 + 2x\right) \\ & = {2x \over 1 + 2x} + c \end{align}

(a)

\begin{align} \int 2x^3 \phantom{0} dx & = 2\int x^3 \phantom{0} dx \\ & = 2 \left( x^4 \over 4 \right) + c \\ & = {x^4 \over 2} + c \\ & = {1 \over 2}x^4 + c \end{align}

(b)

\begin{align} \int (-5) \phantom{0} dx & = -5 \int 1 \phantom{0} dx \\ & = -5(x) + c \\ & = -5x + c \end{align}

(c)

\begin{align} \int \sqrt{x} \phantom{0} dx & = \int x^{1 \over 2} \phantom{0} dx \\ & = {x^{3 \over 2} \over {3 \over 2}} + c \\ & = \left( x^{3 \over 2} \div {3 \over 2} \right) + c \\ & = \left(x^{3 \over 2} \times {2 \over 3} \right) + c \\ & = {2 \over 3}x^{3 \over 2} + c \end{align}

(d)

\begin{align} \int \left(- {1 \over x^2} \right) \phantom{0} dx & = - \int {1 \over x^2} \phantom{0} dx \\ & = - \int x^{-2} \phantom{0} dx \\ & = - \left(x^{-1} \over -1\right) + c \\ & = - (-x^{-1}) + c \\ & = x^{-1} + c \\ & = {1 \over x} + c \end{align}

(e)

\begin{align} \int {1 \over 2x^3} \phantom{0} dx & = {1 \over 2} \int {1 \over x^3} \phantom{0} dx \\ & = {1 \over 2} \int x^{-3} \phantom{0} dx \\ & = {1 \over 2} \left( x^{-2} \over -2 \right) + c \\ & = -{1 \over 4} (x^{-2}) + c \\ & = -{1 \over 4} \left(1 \over x^2 \right) + c \\ & = -{1 \over 4x^2} + c \end{align}

(f)

\begin{align} \int {2 \over \sqrt{x}} \phantom{0} dx & = 2\int {1 \over \sqrt{x}} \phantom{0} dx \\ & = 2 \int x^{-{1 \over 2}} \phantom{0}dx \\ & = 2 \left( x^{1 \over 2} \over {1 \over 2} \right) + c \\ & = 2 \left( x^{1 \over 2} \div {1 \over 2} \right) + c \\ & = 2 \left( x^{1 \over 2} \times 2 \right) + c \\ & = 2 (2x^{1 \over 2}) + c \\ & = 4x^{1 \over 2} + c \\ & = 4\sqrt{x} + c \end{align}

(a)

\begin{align} \int (6x + 3) \phantom{0} dx & = {6x^2 \over 2} + 3x + c \\ & = 3x^2 + 3x + c \end{align}

(b)

\begin{align} \int (3 - \sqrt{x}) \phantom{0} dx & = \int (3 - x^{1 \over 2}) \phantom{0} dx \\ & = 3x - {x^{3 \over 2} \over {3 \over 2}} + c \\ & = 3x - \left( x^{3 \over 2} \div {3 \over 2} \right) + c \\ & = 3x - \left( x^{3 \over 2} \times {2 \over 3} \right) + c \\ & = 3x - {2 \over 3}x^{3 \over 2} + c \end{align}

(c)

\begin{align} \int 3x(x + 2) \phantom{0} dx & = \int 3x^2 + 6x \phantom{0} dx \\ & = {3x^3 \over 3} + {6x^2 \over 2} + c \\ & = x^3 + 3x^2 + c \end{align}

(d)

\begin{align} \int (x - 1)(x + 2) \phantom{0} dx & = \int x^2 + 2x - x - 2 \phantom{0} dx \\ & = \int x^2 + x - 2 \phantom{0} dx \\ & = {x^3 \over 3} + {x^2 \over 2} - 2x + c \\ & = {1 \over 3}x^3 + {1 \over 2}x^2 - 2x + c \end{align}

(e)

\begin{align} \int \sqrt{x}(\sqrt{x} + 3) \phantom{0} dx & = \int x + 3\sqrt{x} \phantom{0} dx \\ & = \int x + 3x^{1 \over 2} \phantom{0} dx \\ & = {x^2 \over 2} + {3x^{3 \over 2} \over {3 \over 2}} + c \\ & = {1 \over 2}x^2 + \left(3x^{3 \over 2} \div {3 \over 2} \right) + c \\ & = {1 \over 2}x^2 + \left(3x^{3 \over 2} \times {2 \over 3} \right) + c \\ & = {1 \over 2}x^2 + 2x^{3 \over 2} + c \end{align}

(f)

\begin{align} \int (2x - \sqrt{x})^2 \phantom{0} dx & = \int (2x)^2 - 2(2x)(\sqrt{x}) + (\sqrt{x})^2 \phantom{0} dx \\ & = \int 4x^2 - 4x^{3 \over 2} + x \phantom{0} dx \\ & = {4x^3 \over 3} - {4x^{5 \over 2} \over {5 \over 2}} + {x^2 \over 2} + c \\ & = {4 \over 3}x^3 - \left(4x^{5 \over 2} \times {2 \over 5} \right) + {1 \over 2}x^2 + c \\ & = {4 \over 3}x^3 - {8 \over 5}x^{5 \over 2} + {1 \over 2}x^2 + c \end{align}

(a)

\begin{align} \require{cancel} \int {2x^2 + 3 \over x^2} \phantom{0} dx & = \int {2\cancel{x^2} \over \cancel{x^2}} + {3 \over x^2} \phantom{0} dx \\ & = \int 2 + 3\left(1 \over x^2\right) \phantom{0} dx \\ & = \int 2 + 3x^{-2} \phantom{0} dx \\ & = 2x + {3x^{-1} \over -1} + c \\ & = 2x - 3x^{-1} + c \\ & = 2x - {3 \over x} + c \end{align}

(b)

\begin{align} \require{cancel} \int {x^2 + 1 \over 2x^2} \phantom{0} dx & = \int {\cancel{x^2} \over 2\cancel{x^2}} + {1 \over 2x^2} \phantom{0} dx \\ & = \int {1 \over 2} + {1 \over 2}\left(1 \over x^2\right) \phantom{0} dx \\ & = \int {1 \over 2} + {1 \over 2} x^{-2} \phantom{0} dx \\ & = {1 \over 2}x + {1 \over 2}\left(x^{-1} \over -1\right) + c \\ & = {1 \over 2}x - {1 \over 2}x^{-1} + c \\ & = {1 \over 2}x - {1 \over 2x} + c \end{align}

(c)

\begin{align} \int {x + 1 \over \sqrt{x}} \phantom{0} dx & = \int {x + 1 \over x^{1 \over 2}} \phantom{0} dx \\ & = \int {x \over x^{1 \over 2}} + {1 \over x^{1 \over 2}} \phantom{0} dx \\ & = \int x^{1 \over 2} + x^{-{1 \over 2}} \phantom{0} dx \\ & = {x^{3 \over 2} \over {3 \over 2}} + {x^{1 \over 2} \over {1 \over 2}} + c \\ & = \left( x^{3 \over 2} \div {3 \over 2} \right) + \left( x^{1 \over 2} \div {1 \over 2} \right) + c \\ & = \left(x^{3 \over 2} \times {2 \over 3}\right) + \left(x^{1 \over 2} \times 2\right) + c \\ & = {2 \over 3}x^{3 \over 2} + 2x^{1 \over 2} + c \\ & = {2 \over 3}x^{3 \over 2} + 2\sqrt{x} + c \end{align}

(a)

\begin{align} \int (3x + 1)^4 \phantom{0} dx & = {(3x + 1)^5 \over (5)(3)} + c \\ & = {(3x + 1)^5 \over 15} + c \\ & = {1 \over 15}(3x + 1)^5 + c \end{align}

(b)

\begin{align} \int (1 - x)^3 \phantom{0} dx & = {(1 - x)^4 \over (4)(-1)} + c \\ & = {(1 - x)^4 \over -4} + c \\ & = -{1 \over 4}(1 - x)^4 + c \end{align}

(c)

\begin{align} \int (2x + 5)^{-3} \phantom{0} dx & = {(2x + 5)^{-2} \over (-2)(2)} + c \\ & = {(2x + 5)^{-2} \over -4} + c \\ & = -{1 \over 4}(2x + 5)^{-2} + c \end{align}

\begin{align} y & = \int {dy \over dx} \phantom{0} dx \\ & = \int 3x^2 - x \phantom{0} dx \\ & = {3x^3 \over 3} - {x^2 \over 2} + c \\ & = x^3 - {1 \over 2}x^2 + c \\ \\ \text{Using } & (2, 4), \\ (4) & = (2)^3 - {1 \over 2}(2)^2 + c \\ 4 & = 8 - 2 + c \\ 4 & = 6 + c \\ - 2 & = c \\ \\ \therefore y & = x^3 - {1 \over 2}x^2 - 2 \end{align}

(a)

\begin{align} {dy \over dx} & = 2x + {3 \over x^2} \\ & = 2x + 3\left(1 \over x^2\right) \\ & = 2x + 3x^{-2} \\ \\ y & = \int {dy \over dx} \phantom{0} dx \\ & = \int 2x + 3x^{-2} \phantom{0} dx \\ & = {2x^2 \over 2} + {3x^{-1} \over -1} + c \\ & = x^2 - 3x^{-1} + c \\ & = x^2 - {3 \over x} + c \\ \\ \text{Using } & (-1, 5), \\ (5) & = (-1)^2 - {3 \over (-1)} + c \\ 5 & = 1 + 3 + c \\ 5 & = 4 + c \\ 1 & = c \\ \\ \therefore y & = x^2 - {3 \over x} + 1 \end{align}

(b)

\begin{align} {dy \over dx} & \propto (x^2 - 1) \\ \\ {dy \over dx} & = k(x^2 - 1) \\ & = kx^2 - k \\ \\ \text{When } & x = 2 \text{ and } {dy \over dx} = 9, \\ (9) & = k(2)^2 - k \\ 9 & = 4k - k \\ 9 & = 3k \\ {9 \over 3} & = k \\ 3 & = k \\ \\ {dy \over dx} & = 3x^2 - 3 \\ \\ y & = \int {dy \over dx} \phantom{0} dx \\ & = \int 3x^2 - 3 \phantom{0} dx \\ & = {3x^3 \over 3} - 3x + c \\ & = x^3 - 3x + c \\ \\ \text{When } & x = 2 \text{ and } y = 3, \\ (3) & = (2)^3 - 3(2) + c \\ 3 & = 8 - 6 + c \\ 3 - 8 + 6 & = c \\ 1 & = c \\ \\ \therefore y & = x^3 - 3x + 1 \\ \\ \text{When } & x = 3, \\ y & = (3)^3 - 3(3) + 1 \\ & = 27 - 9 + 1 \\ & = 19 \end{align}

\begin{align} {d V \over dt} & = 6(2t - 1)^2 + 1 \\ \\ V & = \int {dV \over dt} \phantom{0} dt \\ & = \int 6(2t - 1)^2 + 1 \phantom{0} dt \\ & = {6(2t - 1)^3 \over (3)(2)} + t + c \\ & = {6(2t - 1)^3 \over 6} + t + c \\ & = (2t - 1)^3 + t + c \\ \\ \text{When } & t = 1 \text{ and } V = 4, \\ (4) & = [2(1) - 1]^3 + (1) + c \\ 4 & = 1 + 1 + c \\ 4 & = 2 + c \\ 2 & = c \\ \\ \therefore V & = (2t - 1)^3 + t + 2 \end{align}

(a)

\begin{align} x & = \int {dx \over dt} \phantom{0} dt \\ & = \int 3t^2 + 2 \phantom{0} dt \\ & = {3t^3 \over 3} + 2t + c \\ & = t^3 + 2t + c \\ \\ \text{When } & t = 0 \text{ and } x = 1, \phantom{000000} [\text{Initial radius}] \\ 1 & = (0)^3 + 2(0) + c \\ 1 & = c \\ \\ \therefore x & = t^3 + 2t + 1 \end{align}

(b)

\begin{align} A & = \int {dA \over dt} \phantom{0} dt \\ & = \int 6t^2 - 2t + 1 \phantom{0} dt \\ & = {6t^3 \over 3} - {2t^2 \over 2} + t + c \\ & = 2t^3 - t^2 + t + c \\ \\ \text{When } & t = 2 \text{ and } A = 11, \\ (11) & = 2(2)^3 - (2)^2 + (2) + c \\ 11 & = 16 - 4 + 2 + c \\ 11 & = 14 + c \\ -3 & = c \\ \\ \therefore A & = 2t^3 - t^2 + t - 3 \end{align}

(i)

\begin{align} {d \over dx} \sqrt{6x + 5} & = {d \over dx} (6x + 5)^{1 \over 2} \\ & = {1 \over 2} (6x + 5)^{-{1 \over 2}} . (6) \phantom{000000} [\text{Chain rule}] \\ & = 3(6x + 5)^{-{1 \over 2}} \\ & = {3 \over \sqrt{6x + 5}} \end{align}

(ii)

\begin{align} {d \over dx} \sqrt{6x + 5} & = {3 \over \sqrt{6x + 5}} \\ \\ \implies \int {3 \over \sqrt{6x + 5}} \phantom{.} dx & = \sqrt{6x + 5} \\ \\ \therefore \int {1 \over \sqrt{6x + 5}} \phantom{.} dx & = {1 \over 3} \int {3 \over \sqrt{6x + 5}} \phantom{.} dx \\ & = {1 \over 3} \sqrt{6x + 5} + c \end{align}

(i)

\begin{align} u & = x &&& v & = \sqrt{x + 1} \\ & &&& & = (x + 1)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(x + 1)^{-{1 \over 2}} . (1) \phantom{000000} [\text{Chain rule}] \\ & &&& & = {1 \over 2}(x + 1)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x + 1}} \end{align} \begin{align} {d \over dx} \left(x \sqrt{x + 1} \right) & = (x) \left(1 \over 2\sqrt{x + 1}\right) + \left( \sqrt{x + 1} \right) (1) \phantom{000000} [\text{Product rule}] \\ & = {x \over 2\sqrt{x + 1}} + \sqrt{x + 1} \\ & = {x \over 2\sqrt{x + 1}} + {\sqrt{x + 1} \over 1} \\ & = {x \over 2\sqrt{x + 1}} + {2 \sqrt{x + 1} \left( \sqrt{x + 1} \right) \over 2 \sqrt{x + 1} } \\ & = {x \over 2\sqrt{x + 1}} + { 2(x + 1) \over \sqrt{x + 1}} \\ & = {x + 2(x + 1) \over 2\sqrt{x + 1}} \\ & = {x + 2x + 2 \over 2 \sqrt{x + 1}} \\ & = {3x + 2 \over 2 \sqrt{x + 1}} \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} {d \over dx} \left(x \sqrt{x + 1} \right) & = {3x + 2 \over 2 \sqrt{x + 1}} \\ \\ \implies \int {3x + 2 \over 2 \sqrt{x + 1}} \phantom{.} dx & = x \sqrt{x + 1} \\ \\ \therefore \int {3x + 2 \over \sqrt{x + 1}} \phantom{.} dx & = 2 \int {3x + 2 \over 2 \sqrt{x + 1}} \phantom{.} dx \\ & = 2 \left(x \sqrt{x + 1} \right) \\ & = 2x \sqrt{x + 1} + c \end{align}

(a)

\begin{align} \int \sqrt{6x - 1} \phantom{0} dx & = \int (6x - 1)^{1 \over 2} \phantom{0} dx \\ & = {(6x - 1)^{3 \over 2} \over \left(3 \over 2\right)(6)} \\ & = {(6x - 1)^{3 \over 2} \over 9} \\ & = {1 \over 9}(6x - 1)^{3 \over 2} + c \end{align}

(b)

\begin{align} \int {2 \over (2x - 7)^2} \phantom{0} dx & = 2\int {1 \over (2x - 7)^2} \phantom{0} dx \\ & = 2\int (2x - 7)^{-2} \phantom{0} dx \\ & = 2\left[{(2x - 7)^{-1} \over (-1)(2)}\right] \\ & = {2(2x - 7)^{-1} \over -2} \\ & = -(2x - 7)^{-1} \\ & = -{1 \over 2x - 7} + c \end{align}

(c)

\begin{align} \int {1 \over \sqrt{3 - 2x}} \phantom{0} dx & = \int (3 - 2x)^{-{1 \over 2}} \phantom{0} dx \\ & = {(3 - 2x)^{1 \over 2} \over \left(1 \over 2\right)(-2)} \\ & = {(3 - 2x)^{1 \over 2} \over -1} \\ & = -(3 - 2x)^{1 \over 2} \\ & = -\sqrt{3 - 2x} + c \end{align}

\begin{align} {dy \over dx} & = x^2(x - k) \\ & = x^3 - kx^2 \\ \\ y & = \int {dy \over dx} \phantom{0} dx \\ & = \int (x^3 - kx^2) \phantom{0} dx \\ & = {x^4 \over 4} - {kx^3 \over 3} + c \\ & = {1 \over 4}x^4 - {k \over 3}x^3 + c \\ \\ \text{Using } & (2, -2), \\ (-2) & = {1 \over 4}(2)^4 - {k \over 3}(2)^3 + c \\ -2 & = 4 - {8k \over 3} + c \\ -2 - 4 + {8k \over 3} & = c \\ -6 + {8k \over 3} & = c \\ \\ c & = -6 + {8k \over 3} \phantom{0} \text{ --- (1)} \\ \\ \text{Using } & (4, 2), \\ (2) & = {1 \over 4}(4)^4 - {k \over 3}(4)^3 + c \\ 2 & = 64 - {64k \over 3} + c \\ 2 - 64 & = -{64k \over 3} + c \\ -62 & = -{64k \over 3} + c \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -62 & = -{64k \over 3} + \left(-6 + {8k \over 3}\right) \\ -62 & = -{64k \over 3} -6 + {8k \over 3} \\ -62 + 6 & = -{64k \over 3} + {8k \over 3} \\ -56 & = -{56k \over 3} \\ 56 & = {56k \over 3} \\ 3(56) & = 56k \\ 168 & = 56k \\ {168 \over 56} & = k \\ 3 & = k \\ \\ \text{Substitute } & k = 3 \text{ into (1),} \\ c & = -6 + {8(3) \over 3} \\ & = 2 \\ \\ \therefore y & = {1 \over 4}x^4 - {(3) \over 3}x^3 + (2) \\ y & = {1 \over 4}x^4 - x^3 + 2 \end{align}

\begin{align} {dy \over dx} & \propto [x(2x^2 - 3)] \\ \\ {dy \over dx} & = k[x(2x^2 - 3)] \\ & = kx(2x^2 - 3) \\ & = 2kx^3 - 3kx \\ \\ y & = \int {dy \over dx} \phantom{0} dx \\ & = \int (2kx^3 - 3kx) \phantom{0} dx \\ & = {2kx^4 \over 4} - {3kx^2 \over 2} \\ & = {kx^4 \over 2} - {3k \over 2}x^2 \\ & = {k \over 2}x^4 - {3k \over 2}x^2 + c \\ \\ \text{Using } & (1, 3), \\ (3) & = {k \over 2}(1)^4 - {3k \over 2}(1)^2 + c \\ 3 & = {k \over 2} - {3k \over 2} + c \\ 3 & = -k + c \\ 3 + k & = c \\ \\ c & = k + 3 \phantom{0} \text{ --- (1)} \\ \\ \text{Using } & (2, 9), \\ (9) & = {k \over 2}(2)^4 - {3k \over 2}(2)^2 + c \\ 9 & = 8k - 6k + c \\ 9 & = 2k + c \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 9 & = 2k + (k + 3) \\ 9 & = 2k + k + 3 \\ 9 & = 3k + 3 \\ 9 - 3 & = 3k \\ 6 & = 3k \\ {6 \over 3} & = k \\ 2 & = k \\ \\ \text{Substitute } & k = 2 \text{ into (1),} \\ c & = (2) + 3 \\ & = 5 \\ \\ \therefore y & = {(2) \over 2}x^4 - {3(2) \over 2}x^2 + (5) \\ y & = x^4 - 3x^2 + 5 \end{align}

\begin{align} {dy \over dx} & = x(2 - 3x) \\ & = 2x - 3x^2 \\ \\ y & = \int {dy \over dx} \phantom{0} dx \\ & = \int (2x - 3x^2) \phantom{0} dx \\ & = {2x^2 \over 2} - {3x^3 \over 3} \\ & = x^2 - x^3 + c \\ \\ \text{Using } & (1, 2), \\ (2) & = (1)^2 - (1)^3 + c \\ 2 & = 1 - 1 + c \\ 2 & = c \\ \\ \text{Eqn of curve: } & y = x^2 - x^3 + 2 \\ \\ \text{Using } & (-2, p), \\ (p) & = (-2)^2 - (-2)^3 + 2 \\ p & = 4 - (-8) + 2 \\ & = 14 \end{align}

\begin{align} {d^2y \over dx^2} & = 6x - 16 \\ \\ {dy \over dx} & = \int {d^2y \over dx^2} \phantom{0} dx \\ & = \int (6x - 16) \phantom{0} dx \\ & = {6x^2 \over 2} - 16x \\ & = 3x^2 - 16x + c \\ \\ y & = \int {dy \over dx} \phantom{0} dx \\ & = \int (3x^2 - 16x + c) \phantom{0} dx \\ & = {3x^3 \over 3} - {16x^2 \over 2} + cx + d \\ & = x^3 - 8x^2 + cx + d \\ \\ \text{Using } & (0, 3), \\ (3) & = (0)^3 - 8(0)^2 + c(0) + d \\ 3 & = 0 - 0 + 0 + d \\ 3 & = d \\ \\ \text{Eqn of curve: } & y = x^3 - 8x^2 + cx + 3 \\ \\ \text{Using } & (-1, 0), \\ (0) & = (-1)^3 - 8(-1)^2 + c(-1) + (3) \\ 0 & = -1 - 8 - c + 3 \\ c & = -1 - 8 + 3 \\ & = -6 \\ \\ \therefore y & = x^3 - 8x^2 - 6x + 3 \end{align}

(i)

\begin{align} {d^2y \over dx^2} & = 6x - 4 \\ \\ {dy \over dx} & = \int {d^2y \over dx^2} \phantom{0} dx \\ & = \int (6x - 4) \phantom{0} dx \\ & = {6x^2 \over 2} - 4x \\ & = 3x^2 - 4x + c \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = 0, \phantom{000000} [\text{Minimum point } (1, 5)] \\ 0 & = 3(1)^2 - 4(1) + c \\ 0 & = 3 - 4 + c \\ 0 & = - 1 + c \\ 1 & = c \\ \\ {dy \over dx} & = 3x^2 - 4x + 1 \\ \\ y & = \int {dy \over dx} \phantom{0} dx \\ & = \int (3x^2 - 4x + 1) \phantom{0} dx \\ & = {3x^3 \over 3} - {4x^2 \over 2} + x \\ & = x^3 - 2x^2 + x + d \\ \\ \text{Using } & (1, 5), \\ (5) & = (1)^3 - 2(1)^2 + (1) + d \\ 5 & = 1 - 2 + 1 + d \\ 5 & = 0 + d \\ 5 & = d \\ \\ \therefore y & = x^3 - 2x^2 + x + 5 \end{align}

(ii)

\begin{align} {dy \over dx} & = 3x^2 - 4x + 1 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 3x^2 - 4x + 1 \\ 0 & = (3x - 1)(x - 1) \\ \\ 3x - 1 = 0 \phantom{000} & \text{or} \phantom{000} x - 1 = 0 \\ 3x = 1 \phantom{000} & \phantom{or000-1} x = 1 \text{ [Minimum point } (1, 5)] \\ x = {1 \over 3} \phantom{0()} & \\ \\ \text{Substitute } & x = {1 \over 3} \text{ into eqn of curve,} \\ y & = \left(1 \over 3\right)^3 - 2\left(1 \over 3\right)^2 + \left(1 \over 3\right) + 5 \\ & = {139 \over 27} \\ & = 5{4 \over 27} \\ \\ \text{Substitute } & x = {1 \over 3} \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 6 \left({1 \over 3}\right) - 4 \\ & = -2 \\ \\ \therefore \text{Maximum } & \text{value of } y = 5{4 \over 27} \end{align}

(i)

\begin{align} u & = 2x &&& v & = \sqrt{x + 1} \\ & &&& & = (x + 1)^{1 \over 2} \\ {du \over dx} & = 2 &&& {dv \over dx} & = {1 \over 2}(x + 1)^{-{1 \over 2}} . (1) \phantom{000000} [\text{Chain rule}] \\ & &&& & = {1 \over 2(x + 1)^{1 \over 2}} \end{align} \begin{align} {d \over dx} \left( 2x \over \sqrt{x + 1} \right) & = { (\sqrt{x + 1})(2) - (2x) \left[ 1 \over 2(x + 1)^{1 \over 2} \right] \over (\sqrt{x + 1})^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { 2 \sqrt{x + 1} - {x \over (x + 1)^{1 \over 2}} \over x + 1 } \\ & = { (x + 1)^{1 \over 2} (2 \sqrt{x + 1}) - x \over (x + 1)(x + 1)^{1 \over 2} } \\ & = { 2(x + 1) - x \over (x + 1)^{1 + {1 \over 2}}} \\ & = { 2x + 2 - x \over (x + 1)^{3 \over 2}} \\ & = {x + 2 \over (x + 1)^{3 \over 2}} \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} {d \over dx} \left( 2x \over \sqrt{x + 1} \right) & = {x + 2 \over (x + 1)^{3 \over 2}} \\ \implies \int {x + 2 \over (x + 1)^{3 \over 2}} \phantom{.} dx & = {2x \over \sqrt{x + 1}} \\ \\ \therefore \int {x + 2 \over 4(x + 1)^{3 \over 2}} \phantom{.} dx & = {1 \over 4}\int {x + 2 \over (x + 1)^{3 \over 2}} \phantom{.} dx \\ & = {1 \over 4} \left( {2x \over \sqrt{x + 1}} \right) \\ & = {x \over 2\sqrt{x + 1}} + c \end{align}

(a) Note this fraction is an improper fraction

$$ \require{enclose} \begin{array}{rll} 1 \phantom{000000000}\\ x^2 + 2x + 1 \enclose{longdiv}{ x^2 + 2x \phantom{0000.}}\kern-.2ex \\ -\underline{( x^2 + 2x + 1 ){\phantom{.}}} \\ -1\phantom{0} \end{array} $$
\begin{align} \int {x^2 + 2x \over x^2 + 2x + 1} \phantom{.} dx & = \int 1 - {1 \over x^2 + 2x + 1} \phantom{.} dx \\ & = \int 1 - {1 \over (x + 1)^2} \phantom{.} dx \\ & = \int 1 - (x + 1)^{-2} \phantom{.} dx \\ & = x - {(x + 1)^{-1} \over (-1)(1)} \\ & = x + (x + 1)^{-1} \\ & = x + {1 \over x + 1} + c \end{align}

(b) Note this fraction is an improper fraction

$$ \require{enclose} \begin{array}{rll} {1 \over 3} \phantom{000000000}\\ 9x^2 - 12x + 4 \enclose{longdiv}{ 3x^2 - 4x \phantom{0000.}}\kern-.2ex \\ -\underline{( 3x^2 - 4x + {4 \over 3} ){\phantom{}}} \\ -{4 \over 3} \phantom{.} \end{array} $$
\begin{align} \int {3x^2 - 4x \over 9x^2 - 12x + 4} \phantom{.} dx & = \int {1 \over 3} - { {4 \over 3} \over 9x^2 - 12x + 4} \phantom{.} dx \\ & = \int {1 \over 3} - { {4 \over 3} \over (3x)^2 - 2(3x)(2) + (2)^2} \phantom{.} dx \\ & = \int {1 \over 3} - { {4 \over 3} \over (3x - 2)^2 } \phantom{.} dx \\ & = \int {1 \over 3} - {4 \over 3} (3x - 2)^{-2} \phantom{.} dx \\ & = {1 \over 3}x - {4 \over 3} \left[ (3x - 2)^{-1} \over (-1)(3) \right] \\ & = {1 \over 3}x + {4 \over 9} (3x - 2)^{-1} + c \end{align}

(c)

\begin{align} \int 3a^2 \left( \sqrt{x} + \sqrt[3]{x} \right) \phantom{.} dx & = \int 3a^2 \left( x^{1 \over 2} + x^{1 \over 3} \right) \phantom{.} dx \\ & = 3a^2 \int x^{1 \over 2} + x^{1 \over 3} \phantom{.} dx \\ & = 3a^2 \left( {x^{3 \over 2} \over {3 \over 2}} + {x^{4 \over 3} \over {4 \over 3}} \right) \phantom{.} dx \\ & = 3a^2 \left( {2 \over 3} x^{3 \over 2} + {3 \over 4}x^{4 \over 3} \right) \\ & = 2a^2 x^{3 \over 2} + {9 \over 4} a^2 x^{4 \over 3} + c \end{align}

(i)

\begin{align} {d \over dt} \left[ {1 \over 5} (2t + 1)^{5 \over 2} - {1 \over 3} (2t + 1)^{3 \over 2} \right] & = {1 \over 5}\left(5 \over 2\right)(2t + 1)^{3 \over 2} . (2) - {1 \over 3} \left(3 \over 2\right)(2t + 1)^{1 \over 2} . (2) \phantom{000000} [\text{Chain rule}] \\ & = (2t + 1)^{3 \over 2} - (2t + 1)^{1 \over 2} \\ & = (2t + 1)^{1 \over 2} \left[ (2t + 1) - 1 \right] \\ & = (2t + 1)^{1 \over 2} (2t + 1 - 1) \\ & = (2t + 1)^{1 \over 2} (2t) \\ & = 2t (2t + 1)^{1 \over 2} \end{align}

(ii)

\begin{align} {d \over dt} \left[ {1 \over 5} (2t + 1)^{5 \over 2} - {1 \over 3} (2t + 1)^{3 \over 2} \right] & = 2t (2t + 1)^{1 \over 2} \\ \\ \implies \int 2t(2t + 1)^{1 \over 2} \phantom{.} dt & = {1 \over 5} (2t + 1)^{5 \over 2} - {1 \over 3} (2t + 1)^{3 \over 2} \\ \\ \therefore \int t(2t + 1)^{1 \over 2} \phantom{.} dt & = {1 \over 2} \left[ {1 \over 5} (2t + 1)^{5 \over 2} - {1 \over 3} (2t + 1)^{3 \over 2} \right] \\ & = {1 \over 10} (2t + 1)^{5 \over 2} - {1 \over 6} (2t + 1)^{3 \over 2} + c \end{align}

\begin{align} y & = 5 - x \\ y & = -x + 5 \\ \\ \text{Gradient of tangent} & = -1 \\ \\ {d^2 y \over dx^2} & = 1 - x^2 \\ \\ {dy \over dx} & = \int 1 - x^2 \phantom{.} dx \\ & = x - {x^3 \over 3} + c \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = -1, \\ -1 & = (1) - {(1)^3 \over 3} + c \\ -1 & = 1 - {1 \over 3} + c \\ -1 & = {2 \over 3} + c \\ -{5 \over 3} & = c \\ \\ {dy \over dx} & = x - {x^3 \over 3} - {5 \over 3} \\ \\ y & = \int x - {x^3 \over 3} - {5 \over 3} \phantom{.} dx \\ & = {x^2 \over 2} - {x^4 \over 3(4)} - {5 \over 3}x + c \\ & = {x^2 \over 2} - {x^4 \over 12} - {5 \over 3}x + c \\ \\ \text{Using } & (1, 4), \\ 4 & = {(1)^2 \over 2} - {(1)^4 \over 12} - {5 \over 3}(1) + c \\ 4 & = {1 \over 2} - {1 \over 12} - {5 \over 3} + c \\ 4 & = -{5 \over 4} + c \\ {21 \over 4} & = c \\ \\ \therefore y & = {x^2 \over 2} - {x^4 \over 12} - {5 \over 3}x + {21 \over 4} \end{align}

(i)

\begin{align} {dy \over dx} & = kx + c \\ \\ \text{When } & x = 2 \text{ and } {dy \over dx} = 0, \\ 0 & = k(2) + c \\ 0 & = 2k + c \\ -2k & = c \\ \\ {dy \over dx} & = kx - 2k \\ \\ y & = \int kx - 2k \phantom{.} dx \\ & = {kx^2 \over 2} - 2kx + d \\ \\ \text{Using } & (0, 1), \\ 1 & = {k(0)^2 \over 2} - 2k(0) + d \\ 1 & = 0 - 0 + d \\ 1 & = d \\ \\ y & = {kx^2 \over 2} - 2kx + 1 \\ \\ \text{Using } & (2, -3), \\ -3 & = {k(2)^2 \over 2} - 2k(2) + 1 \\ -3 & = {4k \over 2} - 4k + 1 \\ -3 & = 2k - 4k + 1 \\ -3 & = -2k + 1 \\ 2k & = 1 + 3 \\ 2k & = 4 \\ k & = {4 \over 2} \\ & = 2 \\ \\ \therefore y & = {(2)x^2 \over 2} - 2(2)x + 1 \\ y & = x^2 - 4x + 1 \end{align}

(ii)

\begin{align} \text{From (i), } {dy \over dx} & = kx - 2k \\ & = 2x - 2(2) \\ & = 2x - 4 \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = 2(0) - 4 \\ & = -4 \\ \\ y & = mx + c \\ y & = -4x + c \\ \\ \text{Using } & (0, 1), \\ 1 & = -4(0) + c \\ 1 & = c \\ \\ \text{Eqn of tangent: } & y = -4x + 1 \end{align}

(i)

\begin{align} {d^2 y \over dx^2} & = \lambda x - 18 \\ \\ {dy \over dx} & = {1 \over 2} \lambda x^2 - 18x + c \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = 9, \\ 9 & = {1 \over 2} \lambda (1)^2 - 18(1) + c \\ 9 & = {1 \over 2} \lambda - 18 + c \\ 27 - {1 \over 2} \lambda & = c \\ \\ {dy \over dx} & = {1 \over 2} \lambda x^2 - 18x + 27 - {1 \over 2} \lambda \\ \\ y & = {1 \over 6} \lambda x^3 - 9x^2 + \left(27 - {1 \over 2} \lambda\right)x + d \\ \\ \text{Using } & (0, 0), \\ 0 & = {1 \over 6} \lambda (0)^3 - 9(0)^2 + \left(27 - {1 \over 2} \lambda\right)(0) + d \\ 0 & = 0 - 0 + 0 + d \\ 0 & = d \\ \\ y & = {1 \over 6} \lambda x^3 - 9x^2 + \left(27 - {1 \over 2} \lambda\right)x \\ \\ \text{Using } & (1, 16), \\ 16 & = {1 \over 6} \lambda (1)^3 - 9(1)^2 + \left(27 - {1 \over 2} \lambda\right)(1) \\ 16 & = {1 \over 6} \lambda - 9 + 27 - {1 \over 2} \lambda \\ 16 + 9 - 27 & = {1 \over 6} \lambda - {1 \over 2} \lambda \\ -2 & = -{1 \over 3} \lambda \\ 6 & = \lambda \end{align}

(ii)

\begin{align} y & = {1 \over 6} (6) x^3 - 9x^2 + \left[27 - {1 \over 2} (6)\right]x \\ y & = x^3 - 9x^2 + 24x \end{align}

(iii) Note: A turning point can be either a maximum point or a minimum point. It cannot be a stationary point of inflexion!

\begin{align} {d^2 y \over dx^2} & = \lambda x - 18 \\ & = 6x - 18 \\ \\ \text{From (i), } {dy \over dx} & = {1 \over 2} \lambda x^2 - 18x + 27 - {1 \over 2} \lambda \\ {dy \over dx} & = {1 \over 2} (6) x^2 - 18x + 27 - {1 \over 2}(6) \\ {dy \over dx} & = 3x^2 - 18x + 24 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 3x^2 - 18x + 24 \\ 0 & = x^2 - 6x + 8 \\ 0 & = (x - 2)(x - 4) \end{align} \begin{align} x - 2 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = 2 &&& x & = 4 \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} &&& \text{Substitute } & x = 4 \text{ into eqn of curve,} \\ y & = (2)^3 - 9(2)^2 + 24(2) &&& y & = (4)^3 - 9(4)^2 + 24(4) \\ y & = 20 &&& y & = 16 \\ \\ \text{Substitute } & x = 2 \text{ into } {d^2 y \over dx^2}, &&& \text{Substitute } & x = 4 \text{ into } {d^2 y \over dx^2}, \\ {d^2 y \over dx^2} & = 6(2) - 18 &&& {d^2 y \over dx^2} & = 6(4) - 18 \\ & = -6 < 0 &&& & = 6 > 0 \\ \\ \therefore (2, 20) & \text{ is max. point} &&& \therefore (4, 16) & \text{ is min. point} \end{align}

\begin{align} \text{From Jane's approach, } \int (x + 1) \phantom{.} dx & = {(x + 1)^2 \over 2} + c \\ & = {x^2 + 2x + 1 \over 2} + c \\ & = {x^2 \over 2} + {2x \over 2} + {1 \over 2} + c \\ & = {x^2 \over 2} + x + {1 \over 2} + c \\ & = {x^2 \over 2} + x + c_1, \text{ where } c_1 = {1 \over 2} + c \\ \\ \\ \therefore & \text{ Both are correct} \end{align}

(a)

\begin{align} \int x \phantom{.} dx & = {x^2 \over 2} \\ \\ \int x^2 \phantom{.} dx & = {x^3 \over 3} \\ \\ \\ \text{Consider } \left( \int x \phantom{.} dx \right) \left( \int x \phantom{.} dx \right) & = \left(x^2 \over 2\right)\left(x^2 \over 2\right) \\ & = {x^4 \over 4} \\ & \ne \int x^2 \phantom{.} dx \\ \\ \therefore \text{Not } & \text{true} \end{align}

(b)

\begin{align} \int x \phantom{.} dx & = {x^2 \over 2} \\ \\ \int x^2 \phantom{.} dx & = {x^3 \over 3} \\ \\ \\ \text{Consider } { \int x^2 \phantom{.} dx \over \int x \phantom{.} dx} & = { {x^3 \over 3} \over {x^2 \over 2} } \\ & = {x^3 \over 3} \times {2 \over x^2} \\ & = {2x^3 \over 3x^2} \\ & = {2x \over 3} \\ & = {2 \over 3}x \\ & \ne \int x \phantom{.} dx \\ \\ \therefore \text{Not } & \text{true} \end{align}