Question

Am I doing this right? In the example we modeled the world population in the second half of the 20th century by the equation
P(t) = 2560e^0.017185t.
Use this equation to estimate the average world population during the time period of 1950 to 1980. (Round your answer to the nearest million.)

Answer 1

So I have to integrate from 0 to 30 and divide by 30 right?

Answer 2

\[\frac{ 1 }{ 30 } \int\limits_{0}^{30}2560e^{0.017185t} \]

Answer 3

right

Answer 4

yes

Answer 5

I don't get the answer in millions though

Answer 6

@ganeshie8

Answer 7

tell me if am wrong . that's an exponential function and the
t= time
however am not sure what e is!

Answer 8

yes and e is just a popular number in math. I believe it's irrational. like pi

Answer 9

thats possible. so the function outlook is super confusing. is this the correct function
\[p(t) 2560 \pi (0.017185)^t\]
is that correct? if yes then all we need to do is to substitute the values of t and after that use the slope formula to determine the average.

Answer 10

no, it's
\[P(t)= 2560e^{0.017185t}\]

Answer 11

yeah so did you substitute the values 1950 and 1980 instead of t?

Answer 12

but first, I have to integrate right?, because yes I did. well, 1950= 0 and 1980= 30

Answer 13

so there fore we have the coordinates (1950,0) and (1980,30) use slope formula y2-y1/x2-x1
to find the average.

Answer 14

no, I meant that 1950 represents t=0 and 1980= 30 since 30 years past

Answer 15

I'm pretty sure I have to use integration some how

Answer 16

sure, integrate.
are you confused because of the coefficient in the exponent?

Answer 17

I'm going to show you guys how I do it.

Answer 18

please do

Answer 19

\[\frac{ 1 }{ 30 }*2560\int\limits_{0}^{30}e^{0.017185t} dt\]

Answer 20

\[\frac{ 1 }{ 30 }[2560(\frac{ e^{0.017185t} }{ 0.017185 })]_{0}^{30}\]

Answer 21

Mmmm ok looks good!
Then you just jam a bunch of numbers into your calculator, ya?

Answer 22

lol but I get 8315.14

Answer 23

Hmm I came up with 3349.6

Answer 24

ohh wait a minute I forgo to subtract after plugging in the lower bound zero

Answer 25

\[\large\rm 4965.57 e^{.017185}|_0^{30}\]If you bring all the coefficient junk to the front you have something like this.

Ah yes, e^0 is not 0 :)

Answer 26

yeah now I get 3349.57

Answer 27

So the population averaged 3350 million people over that time period.
Or 3.35 billion people.

Answer 28

ohhh yay :)

Answer 29

thanks!