3.7.5 Pengamiran, SPM Praktis (Kertas 1)

Soalan 11:

\begin{array}{l} Diberi = \int_{2}^{5} g (x) d x = - 2 . Cari \\ (a) nilai bagi \int_{5}^{2} g (x) d x, \\ (b) nilai bagi m jika \int_{2}^{5} [g (x) + m (x)] d x = 19 \end{array}

Penyelesaian:

\begin{array}{l} (a) \int_{5}^{2} g (x) d x = - \int_{2}^{5} g (x) d x \\ = - (- 2) \\ = 2 \end{array}

\begin{array}{l} (b) \int_{2}^{5} [g (x) + m (x)] d x = 19 \\ \int_{2}^{5} g (x) d x + m \int_{2}^{5} x d x = 19 \\ - 2 + m {[\frac{x^{2}}{2}]}_{2}^{5} = 19 \\ \frac{m}{2} {[x^{2}]}_{2}^{5} = 21 \\ \frac{m}{2} [25 - 4] = 21 \\ 21 m = 42 \\ m = 2 \end{array}

Soalan 12:

\begin{array}{l} a) Cari nilai bagi \int_{- 1}^{1} {(3 x + 1)}^{3} d x . \\ (b) Nilaikan \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x . \end{array}

Penyelesaian:

\begin{array}{l} a) {\int_{- 1}^{1} {(3 x + 1)}^{3} d x = [\frac{{(3 x + 1)}^{4}}{4 (3)}]}_{- 1}^{1} \\ = {[\frac{{(3 x + 1)}^{4}}{12}]}_{- 1}^{1} \\ = \frac{1}{12} [4^{4} - {(- 2)}^{4}] \\ = \frac{1}{12} (256 - 16) \\ = 20 \end{array}

\begin{array}{l} (b) \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x = \int_{3}^{4} \frac{1}{{(2 x - 4)}^{\frac{1}{2}}} d x \\ = \int_{3}^{4} {(2 x - 4)}^{- \frac{1}{2}} d x \\ = {[\frac{{(2 x - 4)}^{- \frac{1}{2} + 1}}{\frac{1}{2} (2)}]}_{3}^{4} \\ = {[\sqrt{2 x - 4}]}_{3}^{4} \\ = [\sqrt{2 (4) - 4} - \sqrt{2 (3) - 4}] \\ = 2 - \sqrt{2} \end{array}

Soalan 13:

\begin{array}{l} Diberi y = \frac{x^{2}}{2 x - 1}, tunjukkan \\ \frac{d y}{d x} = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} . Seterusnya, nilaikan \int_{- 2}^{2} \frac{x (x - 1)}{4 {(2 x - 1)}^{2}} d x . \end{array}

Penyelesaian:

\begin{array}{l} y = \frac{x^{2}}{2 x - 1} \\ \frac{d y}{d x} = \frac{(2 x - 1) (2 x) - x (2)}{{(2 x - 1)}^{2}} \\ = \frac{4 x^{2} - 2 x - 2 x^{2}}{{(2 x - 1)}^{2}} \\ = \frac{2 x^{2} - 2 x}{{(2 x - 1)}^{2}} \\ = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} (tertunjuk) \\ \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{8} \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{4} \int_{- 2}^{2} \frac{x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} [(\frac{2^{2}}{2 (2) - 1}) - (\frac{{(- 2)}^{2}}{2 (- 2) - 1})] \\ = \frac{1}{8} [(\frac{4}{3}) - (\frac{4}{- 5})] \\ = \frac{1}{8} (\frac{32}{15}) \\ = \frac{4}{15} \end{array}