3.7.3 Pengamiran, SPM Praktis (Kertas 1)

Soalan 5:

\begin{array}{l} Diberi \int (6 x^{2} + 1) d x = m x^{3} + x + c, \\ dengan keadaan m dan c ialah pemalar, cari \\ (a) nilai m . \\ (b) nilai c jika \int (6 x^{2} + 1) d x = 13 apabila x = 1. \end{array}

Penyelesaian:
(a)

\begin{array}{l} \int (6 x^{2} + 1) d x = m x^{3} + x + c \\ \frac{6 x^{3}}{3} + x + c = m x^{3} + x + c \\ 2 x^{3} + x + c = m x^{3} + x + c \\ Banding kedua-dua belah, \\ Maka, m = 2 \end{array}

(b)

\begin{array}{l} \int (6 x^{2} + 1) d x = 13 apabila x = 1. \\ 2 {(1)}^{3} + 1 + c = 13 \\ 3 + c = 13 \\ c = 10 \end{array}

Soalan 6:

\begin{array}{l} Diberi bahawa \int_{5}^{k} g (x) d x = 6, dan \int_{5}^{k} [g (x) + 2] d x = 14, \\ cari nilai k . \end{array}

Penyelesaian:

\begin{array}{l} \int_{5}^{k} [g (x) + 2] d x = 14 \\ \int_{5}^{k} g (x) d x + \int_{5}^{k} 2 d x = 14 \\ 6 + {[2 x]}_{5}^{k} = 14 \\ 2 (k - 5) = 8 \\ k - 5 = 4 \\ k = 9 \end{array}

Soalan 7:

Diberi \int_{k}^{2} (4 x + 7) d x = 28, hitung nilai yang mungkin bagi k .

Penyelesaian:

\begin{array}{l} \int_{k}^{2} (4 x + 7) d x = 28 \\ {[2 x^{2} + 7 x]}_{k}^{2} = 28 \\ 8 + 14 - (2 k^{2} + 7 k) = 28 \\ 22 - 2 k^{2} - 7 k = 28 \\ 2 k^{2} + 7 k + 6 = 0 \\ (2 k + 3) (k + 2) = 0 \\ k = - \frac{3}{2} atau k = - 2 \end{array}

Soalan 8:

\begin{array}{l} Diberi \int_{- 2}^{3} g (x) d x = 4, dan \int_{- 2}^{3} h (x) d x = 9, cari nilai bagi \\ (a) \int_{- 2}^{3} 5 g (x) d x, \\ (b) m jika \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \end{array}

Penyelesaian:
(a)

\begin{array}{l} \int_{- 2}^{3} 5 g (x) d x = 5 \int_{- 2}^{3} g (x) d x \\ = 5 \times 4 \\ = 20 \end{array}

(b)

\begin{array}{l} \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \\ \int_{- 2}^{3} g (x) d x + 3 \int_{- 2}^{3} h (x) d x + \int_{- 2}^{3} 4 m d x = 12 \\ 4 + 3 (9) + 4 m {[x]}_{- 2}^{3} = 12 \\ 4 m [3 - (- 2)] = - 19 \\ 20 m = - 19 \\ m = - \frac{19}{20} \end{array}