SOAL
Diketahui fungsi force of mortality \(\begin{array}{*{20}{c}} {\mu \left( x \right) = \frac{1}{{x + 1}},}&{x \ge 0} \end{array}\)
Tentukan \(S\left( x \right)\)
- \(\begin{array}{*{20}{c}} {S\left( x \right) = \frac{1}{{x + 1}},}&{x \ge 0} \end{array}\)
- \(\begin{array}{*{20}{c}} {S\left( x \right) = \frac{x}{{x + 1}},}&{x \ge 0} \end{array}\)
- \(\begin{array}{*{20}{c}} {S\left( x \right) = {e^{x + 1}},}&{x \ge 0} \end{array}\)
- \(\begin{array}{*{20}{c}} {S\left( x \right) = \frac{{x + 1}}{x},}&{x \ge 0} \end{array}\)
- \(\begin{array}{*{20}{c}} {S\left( x \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}},}&{x \ge 0} \end{array}\)
| Diketahui | \(\begin{array}{*{20}{c}} {\mu \left( x \right) = \frac{1}{{x + 1}},}&{x \ge 0} \end{array}\) |
| Rumus yang digunakan | \(S\left( x \right) = \exp \left[ { – \int\limits_0^x {\mu \left( t \right)dt} } \right]\) |
| Proses pengerjaan | \(S\left( x \right) = \exp \left[ { – \int\limits_0^x {\left( {\frac{1}{{t + 1}}} \right)dt} } \right] = \exp \left[ { – \ln \left( {x + 1} \right)} \right] = \frac{1}{{x + 1}}\) |
| Jawaban | a. \(\begin{array}{*{20}{c}} {S\left( x \right) = \frac{1}{{x + 1}},}&{x \ge 0} \end{array}\) |