By the definition, f ( t ) = ( t + 10 ) − 2 f(t)=(t+10)-2 f ( t ) = ( t + 10 ) − 2 is the probability density function if
∫ − ∞ + ∞ f ( t ) d t = 1 → ∫ 0 40 ( t + 10 − 2 ) d t = = ∫ 0 40 ( t + 8 ) d t = C ⋅ ( t 2 2 + 8 t ) ∣ 0 40 = C ⋅ ( 1600 2 + 320 ) = 1 → C ⋅ 1120 = 1 ⟶ C = 1 1120 \int\limits_{-\infty}^{+\infty}f(t)dt=1\to \int\limits_0^{40}\left(t+10-2\right)dt=\\[0.3cm]
=\int\limits_0^{40}\left(t+8\right)dt=C\cdot\left.\left(\frac{t^2}{2}+8t\right)\right|_0^{40}=\\[0.3cm]
C\cdot\left(\frac{1600}{2}+320\right)=1\to C\cdot1120=1\longrightarrow\\[0.3cm]
\boxed{C=\frac{1}{1120}} − ∞ ∫ + ∞ f ( t ) d t = 1 → 0 ∫ 40 ( t + 10 − 2 ) d t = = 0 ∫ 40 ( t + 8 ) d t = C ⋅ ( 2 t 2 + 8 t ) ∣ ∣ 0 40 = C ⋅ ( 2 1600 + 320 ) = 1 → C ⋅ 1120 = 1 ⟶ C = 1120 1
Comment: Perhaps the probability distribution function was incorrectly indicated and actually meant f ( t ) = 1 / ( t + 10 ) 2 f(t)=1/(t+10)^2 f ( t ) = 1/ ( t + 10 ) 2 , then
∫ − ∞ + ∞ f ( t ) d t = 1 → ∫ 0 40 ( 1 ( t + 10 ) 2 ) d t = − C ⋅ ( 1 t + 10 ) ∣ 0 40 = − C ⋅ ( 1 50 − 1 10 ) = 1 − C ⋅ 1 − 5 50 = 1 ⟶ C = 50 4 = 25 2 = 12.5 C = 25 2 = 12.5 \int\limits_{-\infty}^{+\infty}f(t)dt=1\to \int\limits_0^{40}\left(\frac{1}{(t+10)^2}\right)dt=\\[0.3cm]
-C\cdot\left.\left(\frac{1}{t+10}\right)\right|_0^{40}=-C\cdot\left(\frac{1}{50}-\frac{1}{10}\right)=1\\[0.3cm] -C\cdot\frac{1-5}{50}=1\longrightarrow C=\frac{50}{4}=\frac{25}{2}=12.5\\[0.3cm]
\boxed{C=\frac{25}{2}=12.5} − ∞ ∫ + ∞ f ( t ) d t = 1 → 0 ∫ 40 ( ( t + 10 ) 2 1 ) d t = − C ⋅ ( t + 10 1 ) ∣ ∣ 0 40 = − C ⋅ ( 50 1 − 10 1 ) = 1 − C ⋅ 50 1 − 5 = 1 ⟶ C = 4 50 = 2 25 = 12.5 C = 2 25 = 12.5
ANSWER
If f ( t ) = ( t + 10 ) − 2 ⟶ C = 1 1120 If f ( t ) = 1 ( t + 10 ) 2 ⟶ C = 25 2 = 12.5 \text{If}\,\,\,f(t)=(t+10)-2\longrightarrow C=\frac{1}{1120}\\[0.3cm]
\text{If}\,\,\,f(t)=\frac{1}{(t+10)^2}\longrightarrow C=\frac{25}{2}=12.5\\[0.3cm] If f ( t ) = ( t + 10 ) − 2 ⟶ C = 1120 1 If f ( t ) = ( t + 10 ) 2 1 ⟶ C = 2 25 = 12.5
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