To estimate the average world population from 1950 to 1980 using the population model provided, we will first express the model clearly:
\[ P(t) = 2560 e^{0.017185t} \]
where \( t \) represents the number of years since 1950, so:
- \( t = 0 \) corresponds to the year 1950.
- \( t = 30 \) corresponds to the year 1980.
To find the average population over the interval from \( t = 0 \) to \( t = 30 \), we can use the formula for the average value of a continuous function over an interval:
\[ \text{Average population} = \frac{1}{b - a} \int_a^b P(t) , dt \]
In our case, \( a = 0 \) and \( b = 30 \):
\[ \text{Average population} = \frac{1}{30 - 0} \int_0^{30} P(t) , dt = \frac{1}{30} \int_0^{30} 2560 e^{0.017185t} , dt \]
Now we can calculate the integral. The integral of \( e^{kt} \) can be computed using the following formula:
\[ \int e^{kt} , dt = \frac{1}{k} e^{kt} + C \]
Applying this to our integral:
\[ \int 2560 e^{0.017185t} , dt = 2560 \cdot \frac{1}{0.017185} e^{0.017185t} + C \]
Now, we evaluate the definite integral:
\[ \int_0^{30} 2560 e^{0.017185t} , dt = 2560 \cdot \frac{1}{0.017185} \left[ e^{0.017185t} \right]_0^{30} \]
Calculating \( e^{0.017185 \cdot 30} \):
\[ e^{0.017185 \cdot 30} \approx e^{0.51555} \approx 1.674 \]
Calculating the lower limit:
\[ e^{0 \cdot 0.017185} = e^0 = 1 \]
Now substituting back into the integral:
\[ = 2560 \cdot \frac{1}{0.017185} (1.674 - 1) \]
Calculating \( 1.674 - 1 = 0.674 \):
\[ = 2560 \cdot \frac{0.674}{0.017185} \]
Calculating \( \frac{0.674}{0.017185} \approx 39.173 \):
\[ = 2560 \cdot 39.173 \approx 100,000.48 \]
Now, substituting this back to find the average population:
\[ \text{Average population} = \frac{100,000.48}{30} \approx 3333.35 \]
Rounding to the nearest million:
\[ \text{Average population} \approx 3333 \text{ million} \text{ (or about } 3.333 \text{ billion)} \]
Thus, the estimated average world population from 1950 to 1980 is approximately 3,333 million.
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