Bab 14 Pengamiran

Posted on May 17, 2016 by user

3.8.3 Pengamiran, SPM Praktis (Kertas 2)

Soalan 5

Dalam rajah di bawah, garis lurus WY ialah normal kepada lengkung

y = \frac{1}{2} x^{2} + 1

pada B (2, 4). Garis lurus BQ adalah selari dengan paksi-y.

Cari
(a) nilai t,
(b) luas rantau yang berlorek,
(c) Isipadu janaan, dalam sebutan π, apabila rantau yang dibatasi oleh lengkung itu, paksi-y dan garis lurus y = 4 dikisarkan melalui 360^o pada paksi-y.

Penyelesaian:

(a)

\begin{array}{l} y = \frac{1}{2} x^{2} + 1 \\ Kecerunan tangen, \frac{d y}{d x} = 2 (\frac{1}{2} x) = x \\ pada titik B, \frac{d y}{d x} = 2 \\ Kecerunan normal, m_{2} = - \frac{1}{2} \\ \frac{4 - 0}{2 - t} = - \frac{1}{2} \\ 8 = - 2 + t \\ t = 10 \end{array}

(b)

Luas rantau yang berlorek

= Luas di bawah lengkung + Luas segi tiga BQY

\begin{array}{l} = \int_{0}^{2} (\frac{1}{2} x^{2} + 1) d x + \frac{1}{2} (10 - 2) (4) \\ = {[\frac{x^{3}}{6} + x]}_{0}^{2} + 16 \\ = [\frac{8}{6} + 2] - 0 + 16 \\ = 19 \frac{1}{3} {unit}^{2} \end{array}

(c)

Pada paksi-y, x = 0, y = ½ (0) + 1 = 1

\begin{array}{l} y = \frac{1}{2} x^{2} + 1 \to x^{2} = 2 y - 2 \\ Isipadu janaan \\ = π \int x^{2} d y \\ = π \int_{1}^{4} (2 y - 2) d y \\ = π {[y^{2} - 2 y]}_{1}^{4} \\ = π [(16 - 8) - (1 - 2)] \\ = 9 π {unit}^{3} \end{array}

Bab 14 Pengamiran

Posted on May 16, 2016 by user

3.7 Pengamiran, SPM Praktis (Kertas 1)

Soalan 2:

Diberi bahawa

\int \frac{4}{{(1 + x)}^{4}} d x = m {(1 + x)}^{n} + c,

Cari nilai-nilai mdan n.

Penyelesaian:

\begin{array}{l} \int \frac{4}{{(1 + x)}^{4}} d x = m {(1 + x)}^{n} + c \\ \int 4 {(1 + x)}^{- 4} d x = m {(1 + x)}^{n} + c \\ \frac{4 {(1 + x)}^{- 3}}{- 3 (1)} + c = m {(1 + x)}^{n} + c \\ - \frac{4}{3} {(1 + x)}^{- 3} + c = m {(1 + x)}^{n} + c \\ m = - \frac{4}{3}, n = - 3 \end{array}

Soalan 3:

Diberi

\int_{- 1}^{2} 2 g (x) d x = 4, dan \int_{- 1}^{2} [m x + 3 g (x)] d x = 15.

Cari nilai pemalar m.

Penyelesaian:

\begin{array}{l} \int_{- 1}^{2} [m x + 3 g (x)] d x = 15 \\ \int_{- 1}^{2} m x d x + \int_{- 1}^{2} 3 g (x) d x = 15 \\ {[\frac{m x^{2}}{2}]}_{- 1}^{2} + 3 \int_{- 1}^{2} g (x) d x = 15 \\ [\frac{m {(2)}^{2}}{2} - \frac{m {(- 1)}^{2}}{2}] + \frac{3}{2} \int_{- 1}^{2} 2 g (x) d x = 15 \\ 2 m - \frac{1}{2} m + \frac{3}{2} (4) = 15 \leftarrow diberi \int_{- 1}^{2} 2 g (x) d x = 4 \\ \frac{3}{2} m + 6 = 15 \\ \frac{3}{2} m = 9 \\ m = 9 \times \frac{2}{3} \\ m = 6 \end{array}

Soalan 4:

Diberi

\frac{d}{d x} (\frac{2 x}{3 - x}) = g (x), cari \int_{1}^{2} g (x) d x .

Penyelesaian:

\begin{array}{l} Diberi \frac{d}{d x} (\frac{2 x}{3 - x}) = g (x) \\ \int g (x) d x = \frac{2 x}{3 - x} \\ dengan itu, \\ \int_{1}^{2} g (x) d x = {[\frac{2 x}{3 - x}]}_{1}^{2} \\ = \frac{2 (2)}{3 - 2} - \frac{2 (1)}{3 - 1} \\ = 4 - 1 \\ = 3 \end{array}

Bab 14 Pengamiran

Posted on May 16, 2016 by user

3.7 Pengamiran, SPM Praktis (Kertas 1)

Soalan 1:

Kamirkan setiap yang berikut terhadap x.

\begin{array}{l} (a) \int (\frac{3}{x^{2}} - \frac{5}{2 x^{3}} + 2) d x \\ (b) \int x^{2} (x^{5} + 2 x) d x \\ (c) \int \frac{3 x^{4} + 2 x}{x^{3}} d x \\ (d) \int \frac{(7 + x) (7 - x)}{x^{4}} d x \\ (e) \int {(5 x - 1)}^{3} d x \\ (f) \int \frac{3}{{(4 x + 7)}^{8}} d x \end{array}

Penyelesaian:

\begin{array}{l} (a) \\ \int (\frac{3}{x^{2}} - \frac{5}{2 x^{3}} + 2) d x \\ = \int (3 x^{- 2} - \frac{5 x^{- 3}}{2} + 2) d x \\ = - 3 x^{- 1} + \frac{5 x^{- 2}}{4} + 2 x + c \\ = - \frac{3}{x} + \frac{5}{4 x^{2}} + 2 x + c \end{array}

\begin{array}{l} (b) \\ \int x^{2} (x^{5} + 2 x) d x \\ = \int (x^{7} + 2 x^{3}) d x \\ = \frac{x^{8}}{8} + \frac{2 x^{4}}{4} + c \\ = \frac{x^{8}}{8} + \frac{x^{4}}{2} + c \end{array}

\begin{array}{l} (c) \\ \int \frac{3 x^{4} + 2 x}{x^{3}} d x \\ = \int (\frac{3 x^{4}}{x^{3}} + \frac{2 x}{x^{3}}) d x = \int (3 x + 2 x^{- 2}) d x \\ = \frac{3 x^{2}}{2} - \frac{2}{x} + c \end{array}

\begin{array}{l} (d) \\ \int \frac{(7 + x) (7 - x)}{x^{4}} d x = \int (\frac{49 - x^{2}}{x^{4}}) d x \\ = \int (\frac{49}{x^{4}} - \frac{1}{x^{2}}) d x = \int (49 x^{- 4} - x^{- 2}) d x \\ = \frac{49 x^{- 3}}{- 3} + \frac{1}{x} + c \\ = - \frac{49}{3 x^{3}} + \frac{1}{x} + c \end{array}

\begin{array}{l} (e) \\ \int {(5 x - 1)}^{3} d x \\ = \frac{{(5 x - 1)}^{4}}{(4) (5)} + c \\ = \frac{1}{20} {(5 x - 1)}^{4} + c \end{array}

\begin{array}{l} (f) \\ \int \frac{3}{{(4 x + 7)}^{8}} d x = \int 3 {(4 x + 7)}^{- 8} d x \\ = \frac{3 {(4 x + 7)}^{- 7}}{(- 7) (4)} + c \\ = - \frac{3}{28 {(4 x + 7)}^{7}} + c \end{array}