(a)

\begin{align} \int_2^5 3x \phantom{0} dx & = \left[ {3x^2 \over 2} \right]_2^5 \\ & = {3(5)^2 \over 2} - {3(2)^2 \over 2} \\ & = 31{1 \over 2} \end{align}

(b)

\begin{align} \int_1^9 x^{1 \over 2} \phantom{0} dx & = \left[ {x^{3 \over 2} \over {3 \over 2}} \right]_1^9 \\ & = { (9)^{3 \over 2} \over {3 \over 2}} - { (1)^{3 \over 2} \over {3 \over 2}} \\ & = 17{1 \over 3} \end{align}

(c)

\begin{align} \int_1^8 {1 \over 2}x^{-{1 \over 3}} \phantom{0} dx & = \left[ {1 \over 2} \left(x^{2 \over 3} \over {2 \over 3}\right) \right]_1^8 \\ & = {1 \over 2} \left[ (8)^{2 \over 3} \over {2 \over 3} \right] - {1 \over 2} \left[ (1)^{2 \over 3} \over {2 \over 3} \right] \\ & = 2{1 \over 4} \end{align}

(d)

\begin{align} \int_2^3 {1 \over 3x^2} \phantom{0} dx & = \int_2^3 {x^{-2} \over 3} \phantom{0} dx \\ & = \left[ {x^{-1} \over 3(-1)} \right]_2^3 \\ & = \left[ - {x^{-1} \over 3} \right]_2^3 \\ & = - {(3)^{-1} \over 3} - \left[ -{(2)^{-1} \over 3}\right] \\ & = {1 \over 18} \end{align}

(e)

\begin{align} \int_4^9 {1 \over \sqrt{x}} \phantom{0} dx & = \int_4^9 x^{-{1 \over 2}} \phantom{0} dx \\ & = \left[ {x^{1 \over 2} \over {1 \over 2}} \right]_4^9 \\ & = { (9)^{1 \over 2} \over {1 \over 2} } - { (4)^{1 \over 2} \over {1 \over 2} } \\ & = 2 \end{align}

(f)

\begin{align} \int_1^4 x\sqrt{x} \phantom{0} dx & = \int_1^4 (x^1)(x^{1 \over 2}) \phantom{0} dx \\ & = \int_1^4 x^{3 \over 2} \phantom{0} dx \phantom{0000000000} [ a^m \times a^n = a^{m + n} ]\\ & = \left[ {x^{5 \over 2} \over {5 \over 2}} \right]_1^4 \\ & = { (4)^{5 \over 2} \over {5 \over 2} } - { (1)^{5 \over 2} \over {5 \over 2} } \\ & = 12{2 \over 5} \end{align}

(a)

\begin{align} \int_{-1}^1 (8x - 4) \phantom{0} dx & = \left[ {8x^2 \over 2} - 4x \right]_{-1}^1 \\ & = \left[4(1)^2 - 4(1) \right] - \left[4(-1)^2 - 4(-1)\right] \\ & = -8 \end{align}

(b)

\begin{align} \int_{-1}^0 (3x^2 - 2x + 5) \phantom{0} dx & = \left[ {3x^3 \over 3} - {2x^2 \over 2} + 5x \right]_{-1}^0 \\ & = \left[ x^3 - x^2 + 5x \right]_{-1}^0 \\ & = \left[ (0)^3 - (0)^2 + 5(0) \right] - \left[ (-1)^3 - (-1)^2 + 5(-1) \right] \\ & = 7 \end{align}

(c)

\begin{align} \int_1^4 (6x - 3\sqrt{x}) \phantom{0} dx & = \int_1^4 (6x - 3x^{1 \over 2}) \phantom{0} dx \\ & = \left[ {6x^2 \over 2} - {3x^{3 \over 2} \over {3 \over 2}} \right]_1^4 \\ & = \left[ 3x^2 - 2x^{3 \over 2} \right]_1^4 \\ & = \left[ 3(4)^2 - 2(4)^{3 \over 2} \right] - \left[ 3(1)^2 - 2(1)^{3 \over 2} \right] \\ & = 31 \end{align}

(a)

\begin{align} \int_0^2 x(x^2 - 2) \phantom{0} dx & = \int_0^2 x^3 - 2x \phantom{0} dx \\ & = \left[ {x^4 \over 4} - {2x^2 \over 2} \right]_0^2 \\ & = \left[ {1 \over 4}x^4 - x^2 \right]_0^2 \\ & = \left[ {1 \over 4}(2)^4 - (2)^2\right] - \left[ {1 \over 4}(0)^4 - (0)^2\right] \\ & = 0 \end{align}

(b)

\begin{align} \int_1^2 (x + 1)(x - 2) \phantom{0} dx & = \int_1^2 x^2 - 2x + x - 2 \phantom{0} dx \\ & = \int_1^2 x^2 - x - 2 \phantom{0} dx \\ & = \left[ {x^3 \over 3} - {x^2 \over 2} - 2x \right]_1^2 \\ & = \left[ {1 \over 3}x^3 - {1 \over 2}x^2 - 2x \right]_1^2 \\ & = \left[ {1 \over 3}(2)^3 - {1 \over 2}(2)^2 - 2(2) \right] - \left[ {1 \over 3}(1)^3 - {1 \over 2}(1)^2 - 2(1)\right] \\ & = -1{1 \over 6} \end{align}

(a)

\begin{align} \int_1^4 {x^2 + 1 \over x^2} \phantom{0} dx & = \int_1^4 {x^2 \over x^2} + {1 \over x^2} \phantom{0} dx \\ & = \int_1^4 1 + x^{-2} \phantom{0} dx \\ & = \left[ x + {x^{-1} \over -1 } \right]_1^4 \\ & = \left[ x - x^{-1} \right]_1^4 \\ & = \left[ x - {1 \over x} \right]_1^4 \\ & = \left[ (4) - {1 \over (4)} \right] - \left[ (1) - {1 \over (1)} \right] \\ & = 3{3 \over 4} \end{align}

(b)

\begin{align} \int_1^2 {1 - 2t^3 \over t^2} \phantom{0} dt & = \int_1^2 {1 \over t^2} - {2t^3 \over t^2} \phantom{0} dt \\ & = \int_1^2 t^{-2} - 2t \phantom{0} dt \\ & = \left[ {t^{-1} \over -1} - {2t^2 \over 2} \right]_1^2 \\ & = \left[ -t^{-1} - t^2 \right]_1^2 \\ & = \left[-{1 \over t} - t^2 \right]_1^2 \\ & = \left[-{1 \over (2)} - (2)^2 \right] - \left[-{1 \over (1)} - (1)^2 \right] \\ & = -2{1 \over 2} \end{align}

(c)

\begin{align} \int_1^4 {2x - 1 \over \sqrt{x}} \phantom{0} dx & = \int_1^4 {2x - 1 \over x^{1 \over 2}} \phantom{0} dx \\ & = \int_1^4 {2x \over x^{1 \over 2}} - {1 \over x^{1 \over 2}} \phantom{0} dx \\ & = \int_1^4 2x^{1 \over 2} - x^{-{1 \over 2}} \phantom{0} dx \\ & = \left[ {2x^{3 \over 2} \over {3 \over 2}} - {x^{1 \over 2} \over {1 \over 2}} \right]_1^4 \\ & = \left[ {4x^{3 \over 2} \over 3} - 2x^{1 \over 2} \right]_1^4 \\ & = \left[ {4(4)^{3 \over 2} \over 3} - 2(4)^{1 \over 2} \right] - \left[ {4(1)^{3 \over 2} \over 3} - 2(1)^{1 \over 2} \right] \\ & = 7{1 \over 3} \end{align}

(d)

\begin{align} \int_1^9 {3 - 2\sqrt{x} \over x^2} \phantom{0} dx & = \int_1^9 {3 - 2x^{1 \over 2} \over x^2} \phantom{0} dx \\ & = \int_1^9 {3 \over x^2} - {2x^{1 \over 2} \over x^2} \phantom{0} dx \\ & = \int_1^9 3x^{-2} - 2x^{-{3 \over 2}} \phantom{0} dx \\ & = \left[ {3x^{-1} \over -1} - {2x^{-{1 \over 2}} \over -{1 \over 2}} \right]_1^9 \\ & = \left[ -3x^{-1} + 4x^{-{1 \over 2}} \right]_1^9 \\ & = \left[ -{3 \over x} + {4 \over \sqrt{x}} \right]_1^9 \\ & = \left[ -{3 \over (9)} + {4 \over \sqrt{9}} \right] - \left[ -{3 \over (1)} + {4 \over \sqrt{1}} \right] \\ & = 0 \end{align}

(a)

\begin{align} \int_1^4 \left(\sqrt{x} - {2 \over \sqrt{x}} \right) \phantom{0} dx & = \int_1^4 x^{1 \over 2} - 2x^{-{1 \over 2}} \phantom{0} dx \\ & = \left[ {x^{3 \over 2} \over {3 \over 2}} - {2x^{1 \over 2} \over {1 \over 2}} \right]_1^4 \\ & = \left[ {2x^{3 \over 2} \over 3} - 4x^{1 \over 2} \right]_1^4 \\ & = \left[ {2(4)^{3 \over 2} \over 3} - 4(4)^{1 \over 2} \right] - \left[ {2(1)^{3 \over 2} \over 3} - 4(1)^{1 \over 2} \right] \\ & = {2 \over 3} \end{align}

(b)

\begin{align} \int_1^2 \left(x^2 - {4 \over x^2} \right) \phantom{0} dx & = \int_1^2 x^2 - 4x^{-2} \phantom{0} dx \\ & = \left[ {x^3 \over 3} - {4x^{-1} \over -1} \right]_1^2 \\ & = \left[ {1 \over 3}x^3 + 4x^{-1} \right]_1^2 \\ & = \left[ {1 \over 3}x^3 + {4 \over x} \right]_1^2 \\ & = \left[ {1 \over 3}(2)^3 + {4 \over (2)} \right] - \left[ {1 \over 3}(1)^3 + {4 \over (1)} \right] \\ & = {1 \over 3} \end{align}

(c)

\begin{align} \int_1^2 \left(8x^3 - 2 + {1 \over 2x^2} \right) \phantom{0} dx & = \int_1^2 8x^3 - 2 + {1 \over 2}x^{-2} \phantom{0} dx \\ & = \left[ {8x^4 \over 4} - 2x + {1 \over 2}\left(x^{-1} \over -1\right) \right]_1^2 \\ & = \left[ 2x^4 - 2x - {1 \over 2} x^{-1} \right]_1^2 \\ & = \left[ 2x^4 - 2x - {1 \over 2x} \right]_1^2 \\ & = \left[ 2(2)^4 - 2(2) - {1 \over 2(2)} \right] - \left[ 2(1)^4 - 2(1) - {1 \over 2(1)} \right] \\ & = 28{1 \over 4} \end{align}

(a)

\begin{align} \int_{-1}^0 x(x - a)(x + a) \phantom{0} dx & = \int_{-1}^0 x(x^2 - a^2) \phantom{0} dx \phantom{000000} [\text{Identity: } (a + b)(a - b) = a^2 - b^2] \\ & = \int_{-1}^0 x^3 - a^2x \phantom{0} dx \\ & = \left[ {x^4 \over 4} - {a^2x^2 \over 2} \right]_{-1}^0 \\ & = \left[ {1 \over 4}x^4 - {a^2 \over 2}x^2 \right]_{-1}^0 \\ & = \left[ {1 \over 4}(0)^4 - {a^2 \over 2}(0)^2 \right] - \left[ {1 \over 4}(-1)^4 - {a^2 \over 2}(-1)^2 \right] \\ & = \left[ {1 \over 4}(0) - {a^2 \over 2}(0) \right] - \left[ {1 \over 4}(1) - {a^2 \over 2}(1) \right] \\ & = 0 - \left[ {1 \over 4} - {a^2 \over 2} \right] \\ & = -{1 \over 4} + {a^2 \over 2} \\ & = {a^2 \over 2} - {1 \over 4} \end{align}

(b)

\begin{align} \int_0^4 \sqrt{x}(a - \sqrt{x}) \phantom{0} dx & = \int_0^4 a\sqrt{x} - x \phantom{0} dx \\ & = \int_0^4 ax^{1 \over 2} - x \phantom{0} dx \\ & = \left[ {ax^{3 \over 2} \over {3 \over 2}} - {x^2 \over 2} \right]_0^4 \\ & = \left[ {2a \over 3}x^{3 \over 2} - {1 \over 2}x^2 \right]_0^4 \\ & = \left[ {2a \over 3}(4)^{3 \over 2} - {1 \over 2}(4)^2 \right] - \left[ {2a \over 3}(0)^{3 \over 2} - {1 \over 2}(0)^2 \right] \\ & = \left[ {2a \over 3}(8) - {1 \over 2}(16) \right] - \left[ {2a \over 3}(0) - {1 \over 2}(0) \right] \\ & = \left[ {16a \over 3} - 8 \right] - 0 \\ & = {16 \over 3}a - 8 \end{align}

(c)

\begin{align} \int_1^2 {2t^2 + a \over t^2} \phantom{0} dt & = \int_1^2 {2t^2 \over t^2} + {a \over t^2} \phantom{0} dt \\ & = \int_1^2 2 + at^{-2} \phantom{0} dt \\ & = \left[ 2t + {at^{-1} \over -1} \right]_1^2 \\ & = \left[ 2t - at^{-1} \right]_1^2 \\ & = \left[ 2t - {a \over t} \right]_1^2 \\ & = \left[ 2(2) - {a \over (2)} \right] - \left[ 2(1) - {a \over (1)} \right] \\ & = \left[ 4 - {a \over 2} \right] - \left[ 2 - a \right] \\ & = 4 - {a \over 2} - 2 + a \\ & = 2 + {a \over 2} \end{align}

(a)

\begin{align} \int_1^a 4x \phantom{0} dx & = \left[ {4x^2 \over 2} \right]_1^a \\ & = \left[ 2x^2 \right]_1^a \\ & = \left[ 2(a)^2 \right] - \left[ 2(1)^2 \right] \\ & = 2a^2 - 2 \\ \\ \int_2^1 (6x - 3x^2) \phantom{0} dx & = \left[ {6x^2 \over 2} - {3x^3 \over 3} \right]_2^1 \\ & = \left[ 3x^2 - x^3 \right]_2^1 \\ & = \left[ 3(1)^2 - (1)^3 \right] - \left[ 3(2)^2 - (2)^3 \right] \\ & = -2 \\ \\ \text{Since } \int_1^a 4x \phantom{0} dx & = \int_2^1 (6x - 3x^2) \phantom{0} dx, \\ 2a^2 - 2 & = - 2 \\ 2a^2 & = -2 + 2 \\ 2a^2 & = 0 \\ a^2 & = 0 \\ a & = \pm\sqrt{0} \\ & = \pm 0 \\ & = 0 \end{align}

(b)

\begin{align} \int_{-2}^k 3(2 + x)^2 \phantom{0} dx & = \left[{3(2 + x)^3 \over (3)(1)}\right]_{-2}^k \\ & = \left[ (2 + x)^3 \right]_{-2}^k \\ & = [2 + (k)]^3 - [2 + (-2)]^3 \\ & = (2 + k)^3 - 0^3 \\ & = (2 + k)^3 \\ \\ \text{Since } \int_{-2}^k 3(2 + x)^2 \phantom{0} dx & = 64, \\ (2 + k)^3 & = 64 \\ 2 + k & = \sqrt[3]{64} \\ 2 + k & = 4 \\ k & = 4 - 2 \\ & = 2 \end{align}

\begin{align} & \phantom{0 /} \int_0^1 {(2x - 3)^4 - 7 \over 2(2x - 3)^2} \phantom{0} dx \\ & = \int_0^1 {(2x - 3)^4 \over 2(2x - 3)^2} - {7 \over 2(2x - 3)^2} \phantom{0} dx \\ & = \int_0^1 {1 \over 2}(2x - 3)^2 - {7 \over 2}(2x - 3)^{-2} \phantom{0} dx \\ & = \left\{ {1 \over 2} \left[ (2x - 3)^3 \over (3)(2) \right] - {7 \over 2} \left[ (2x - 3)^{-1} \over (-1)(2) \right] \right\}_0^1 \\ & = \left[ {1 \over 12} (2x - 3)^3 + {7 \over 4} (2x - 3)^{-1} \right]_0^1 \\ & = \left[ {1 \over 12} [2(1) - 3)]^3 + {7 \over 4} [2(1) - 3]^{-1} \right] - \left[ {1 \over 12} [2(0) - 3)]^3 + {7 \over 4} [2(0) - 3]^{-1} \right] \\ & = -1{5 \over 6} - \left(- {34 \over 12} \right) \\ & = 1 \end{align}

(i)

\begin{align} u & = x &&& v & = \sqrt{1 + 2x^2} \\ & &&& & = (1 + 2x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(1 + 2x^2)^{-{1 \over 2}} . (4x) \\ & &&& & = 2x (1 + 2x^2)^{-{1 \over 2}} \\ & &&& & = {2x \over \sqrt{1 + 2x^2}} \end{align} \begin{align} {dy \over dx} & = (x)\left(2x \over \sqrt{1 + 2x^2}\right) + (\sqrt{1 + 2x^2})(1) \phantom{000000} [\text{Product rule}] \\ & = {x \over 1} \left(2x \over \sqrt{1 + 2x^2}\right) + \sqrt{1 + 2x^2} \\ & = {2x^2 \over \sqrt{1 + 2x^2} } + { \sqrt{1 + 2x^2} \over 1} \\ & = {2x^2 \over \sqrt{1 + 2x^2} } + { \sqrt{1 + 2x^2} (\sqrt{1 + 2x^2}) \over \sqrt{1 + 2x^2} } \\ & = {2x^2 \over \sqrt{1 + 2x^2} } + { 1 + 2x^2 \over \sqrt{1 + 2x^2} } \\ & = {1 + 4x^2 \over \sqrt{1 + 2x^2} } \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} {dy \over dx} & = {1 + 4x^2 \over \sqrt{1 + 2x^2}} \\ \\ \implies \int {1 + 4x^2 \over \sqrt{1 + 2x^2}} \phantom{.} dx & = x \sqrt{1 + 2x^2} \\ \\ \therefore \int_0^2 {1 + 4x^2 \over \sqrt{1 + 2x^2}} \phantom{0} dx & = \left[ x\sqrt{1 + 2x^2} \right]_0^2 \\ & = \left[ (2)\sqrt{1 + 2(2)^2} \right] - \left[ (0)\sqrt{1 + 2(0)^2} \right] \\ & = \left[ 2\sqrt{9} \right] - 0 \\ & = 6 \end{align}

(i)

\begin{align} \text{L.H.S} & = (x - 4)(x^2 + 4x + 16) \\ & = x^3 + 4x^2 + 16x - 4x^2 - 16x - 64 \\ & = x^3 + 4x^2 - 4x^2 + 16x - 16x - 64 \\ & = x^3 - 64 \\ & = \text{R.H.S} \end{align}

(ii)

\begin{align} \require{cancel} \int_{-1}^3 {x^3 - 64 \over x - 4} \phantom{0} dx & = \int_{-1}^3 {\cancel{(x - 4)}(x^2 + 4x + 16) \over \cancel{x - 4}} \phantom{0} dx \phantom{000000} [\text{Apply result from (i)}] \\ & = \int_{-1}^3 x^2 + 4x + 16 \phantom{0} dx \\ & = \left[ {x^3 \over 3} + {4x^2 \over 2} + 16x \right]_{-1}^3 \\ & = \left[ {(3)^3 \over 3} + {4(3)^2 \over 2} + 16(3) \right] - \left[ {(-1)^3 \over 3} + {4(-1)^2 \over 2} + 16(-1) \right] \\ & = 75 - \left(-14{1 \over 3}\right) \\ & = 89{1 \over 3} \end{align}

(i)

\begin{align} \int_0^2 x \phantom{0} dx & = \left[ {x^2 \over 2} \right]_0^2 \\ & = \left[ {(2)^2 \over 2} \right] - \left[ {(0)^2 \over 2} \right] \\ & = 2 - 0 \\ & = 2 \end{align}

(ii) Consider the function $y = x$.

• For $ 0 < x < 2 $, the function is positive.
• For $ -2 < x < 0 $, the function is negative. So the modulus of the function becomes $y = - x$.

\begin{align} \int_{-2}^2 |x| \phantom{0} dx & = \int_{-2}^0 |x| \phantom{0} dx + \int_0^2 |x| \phantom{0} dx \\ & = \int_{-2}^0 -x \phantom{0} dx + \int_0^2 x \phantom{0} dx \\ & = \left[ -{x^2 \over 2} \right]_{-2}^0 + 2 \\ & = \left[ -{(0)^2 \over 2} \right] - \left[ -{(-2)^2 \over 2} \right] + 2 \\ & = 0 - \left[ -{4 \over 2} \right] + 2 \\ & = 4 \end{align}

(i)

\begin{align} C & = 8000 \left(30 + 3\int_0^x t^{1 \over 4} \phantom{0} dt \right) \\ & = 240 \phantom{.} 000 + 24 \phantom{.} 000 \int_0^x t^{1 \over 4} \phantom{0} dt \\ & = 24 \phantom{.} 0000 + 24 \phantom{.} 000 \left[ {t^{5 \over 4} \over {5 \over 4}} \right]_0^x \\ & = 240 \phantom{.} 000 + 24 \phantom{.} 000 \left[ {4t^{5 \over 4} \over 5} \right]_0^x \\ & = 240 \phantom{.} 000 + 24 \phantom{.} 000 \left[ {4x^{5 \over 4} \over 5} - {4(0)^{5 \over 4} \over 5} \right] \\ & = 240 \phantom{.} 000 + 24 \phantom{.} 000 \left[ {4x^{5 \over 4} \over 5} - 0 \right] \\ & = 240 \phantom{.} 000 + 24 \phantom{.} 000 \left( {4x^{5 \over 4} \over 5}\right) \\ & = 240 \phantom{.} 000 + 24 \phantom{.} 000 \left( {4 \over 5}x^{5 \over 4} \right) \\ & = 240 \phantom{.} 000 + 19 \phantom{.} 200x^{5 \over 4} \\ \\ \text{When } & x = 1, \\ C & = 240 \phantom{.} 000 + 19 \phantom{.} 200 (1)^{5 \over 4} \\ & = 240 \phantom{.} 000 + 19 \phantom{.} 200(1) \\ & = \$259 \phantom{.} 200 \end{align}

(ii)

\begin{align} C & = 240 \phantom{.} 000 + 19 \phantom{.} 200x^{5 \over 4} \\ \\ \text{When } & x = 3, \\ C & = 240 \phantom{.} 000 + 19 \phantom{.} 200 (3)^{5 \over 4} \\ & = 315 \phantom{.} 805.86 \\ & \approx \$316 \phantom{.} 000 \end{align}

(iii)

\begin{align} C & = 240 \phantom{.} 000 + 19 \phantom{.} 200x^{5 \over 4} \\ \\ \text{When } & x = 5, \\ C & = 240 \phantom{.} 000 + 19 \phantom{.} 200 (5)^{5 \over 4} \\ & = 383 \phantom{.} 553.48 \\ & \approx \$384 \phantom{.} 000 \end{align}

\begin{align} V & = \int_0^4 {dV \over dt} \phantom{0} dt \\ & = \int_0^4 20 \phantom{.} 000(t - 6) \phantom{0} dt \\ & = \int_0^4 (20 \phantom{.} 000t - 120 \phantom{.} 000) \phantom{0} dt \\ & = \left[ {20 \phantom{.} 000t^2 \over 2} - 120 \phantom{.} 000t \right]_0^4 \\ & = \left[ 10 \phantom{.} 000t^2 - 120 \phantom{.} 000t \right]_0^4 \\ & = \left[ 10 \phantom{.} 000(4)^2 - 120 \phantom{.} 000(4) \right] - \left[ 10 \phantom{.} 000(0)^2 - 120 \phantom{.} 000(0) \right] \\ & = -320 \phantom{.} 000 - 0 \\ & = -320 \phantom{.} 000 \\ \\ \therefore \text{Total} & \text{ loss of value} = \$ 320 \phantom{.} 000 \end{align}

(i)

\begin{align} \int_0^8 f(x) \phantom{0} dx & = \int_0^3 f(x) \phantom{0} dx + \int_3^8 f(x) \phantom{0} dx \\ & = 12 + 12 \\ & = 24 \end{align}

(ii)

\begin{align} \int_3^8 [ 1 + 2f(x)] \phantom{0} dx & = \int_3^8 1 \phantom{0} dx + 2\int_3^8 f(x) \phantom{0} dx \\ & = \left[ \phantom{.} x \phantom{.} \right]_3^8 + 2(12) \\ & = \left[8 - 3\right] + 24 \\ & = 5 + 24 \\ & = 29 \end{align}

(iii)

\begin{align} \int_3^8 2f(x) \phantom{0} dx + 1 & = 2\int_3^8 f(x) \phantom{0} dx + 1 \\ & = 2(12) + 1 \\ & = 24 + 1 \\ & = 25 \end{align}

Further part

\begin{align} \int_3^8 [f(x) - mx] \phantom{0} dx & = 0 \\ \int_3^8 f(x) \phantom{0} dx - \int_3^8 mx \phantom{0} dx & = 0 \\ 12 - \left[ {mx^2 \over 2} \right]_3^8 & = 0 \\ 12 - \left[ {m(8)^2 \over 2} - {m(3)^2 \over 2} \right] & = 0 \\ 12 - \left[ {64m \over 2} - {9m \over 2} \right] & = 0 \\ 12 - {55m \over 2} & = 0 \\ 12 & = {55m \over 2} \\ 2(12) & = 55m \\ 24 & = 55m \\ {24 \over 55} & = m \end{align}

(i)

\begin{align} \int_{-3}^{-1} f(x) \phantom{.} dx + \int_{-1}^{4} f(x) \phantom{.} dx & = \int_{-3}^{4} f(x) \phantom{.} dx \\ \int_{-3}^{-1} f(x) \phantom{.} dx + b & = a \\ \int_{-3}^{-1} f(x) \phantom{.} dx & = a - b \end{align}

(ii)

\begin{align} \int_{-1}^{-3} 2f(x) \phantom{.} dx & = - \int_{-3}^{-1} 2f(x) \phantom{.} dx \\ & = - 2 \int_{-3}^{-1} f(x) \phantom{.} dx \\ & = - 2( a - b) \\ & = -2a + 2b \end{align}

(i)

\begin{align} y & = x^2 \\ \\ \text{When } & x = -2, \\ y & = (-2)^2 \\ y & = 4 \\ \\ \text{Left limit: } & (-2, 4) \\ \\ \\ \text{When } & x = 2, \\ y & = (2)^2 \\ y & = 4 \\ \\ \text{Right limit: } & (2, 4) \\ \\ \\ \text{When } & x = 0, \\ y & = 0 \\ \\ \text{Intercept: } & (0, 0) \end{align}

(ii)(a) Referring to the sketch in (i), the curve is symmetrical about the y-axis

\begin{align} \int_0^1 x^2 \phantom{.} dx & = \int_{-1}^0 x^2 \phantom{.} dx \\ & = - \int_0^{-1} x^2 \phantom{.} dx \\ & = -m \end{align}

(ii)(b)

\begin{align} \int_{-1}^1 x^2 \phantom{.} dx & = 2 \int_0^1 x^2 \phantom{.} dx \\ & = 2 (-m) \\ & = -2m \end{align}

(ii)(c)

\begin{align} \int_0^{-1} -x^2 \phantom{.} dx & = - \int_{-1}^0 - x^2 \phantom{.} dx \\ & = \int_{-1}^0 x^2 \phantom{.} dx \\ & = -m \end{align}