3.2 Pengamiran Melalui Penggantian - Matematik Tambahan SPM

3.2 Pengamiran Melalui Penggantian

1. Diberi bahawa

{\int (a x + b)}^{n} d x, n \neq - 1.

(A) Kaedah Penggantian,

\begin{array}{l} Katakan u = a x + b \\ Oleh itu, \frac{d u}{d x} = a \\ ∴ d x = \frac{d u}{a} \end{array}

Soalan 1:

Cari {\int (3 x + 5)}^{3} d x .

Penyelesaian:

\begin{array}{l} Katakan u = 3 x + 5 \\ \frac{d u}{d x} = 3 \\ d x = \frac{d u}{3} \\ {\int (3 x + 5)}^{3} d x \\ = \int u^{3} \frac{d u}{3} \leftarrow \begin{array}{l} gantikan 3 x + 5 = u \\ dan d x = \frac{d u}{3} \end{array} \\ = \frac{1}{3} \int u^{3} d u \\ = \frac{1}{3} (\frac{u^{4}}{4}) + c \\ = \frac{1}{3} (\frac{{(3 x + 5)}^{4}}{4}) + c \leftarrow \begin{array}{l} ganti balik \\ u = 3 x + 5 \end{array} \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}

(B) Kaedah Rumus

\begin{array}{l} \int {(a x + b)}^{n} = \frac{{(a x + b)}^{n + 1}}{(n + 1) a} + c \\ Oleh itu, \\ \int {(3 x + 5)}^{3} d x = \frac{{(3 x + 5)}^{4}}{4 (3)} + c \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}

Soalan 2:

Cari,

\begin{array}{l} (a) \int \frac{2}{7} {(5 - x)}^{- 4} d x \\ (b) \int \frac{2}{3 {(9 x - 2)}^{5}} d x \end{array}

Penyelesaian:

\begin{array}{l} (a) \int \frac{2}{7} {(5 - x)}^{- 4} d x = \frac{2 {(5 - x)}^{- 3}}{7 (- 3) (- 1)} + c \\ = \frac{2}{21 {(5 - x)}^{3}} + c \end{array}

\begin{array}{l} (b) \int \frac{2}{3 {(9 x - 2)}^{5}} d x = \int \frac{2 {(9 x - 2)}^{- 5}}{3} d x \\ = \frac{2 {(9 x - 2)}^{- 4}}{3 (- 4) (9)} + c \\ = - \frac{2}{108 {(9 x - 2)}^{4}} + c \\ = - \frac{1}{54 {(9 x - 2)}^{4}} + c \end{array}